Thanks again for your help and your advice, Dr. Charles Elliott (correct me if I'm wrong). I will try to digest everything. Hopefully.

Hello again, sorry for the trouble. I've tried NDSolve method to search for "z" instead of ParametricNDSolveValue So here's what I done but still there's some error.

prog := Module[{Twater = 300., Tnew = 0, Cwater, Dwater, Ewater, 
    ClassicalRungeKuttaCoefficients, f, x, y, z = 60, point, ans, 
    tol = 0.001, iters = 0, iterLim = 100, prevTnew = 10^5, 
    plotList = {}}, $MinPrecison = 50;
   Cwater = 
    QuantityMagnitude[
     ThermodynamicData["Water", 
      "ThermalConductivity", {"Temperature" -> 
        Quantity[Twater, "Kelvins"]}]];
   ans = "Twinkle, twinkle, little star,
             How I wonder what you are.
             Up above the world so high,
             Like a diamond in the sky.
             Twinkle, twinkle, little star,
             How I wonder what you are!";

   While[(Abs[prevTnew - Tnew] > tol) && (iters++ < iterLim), 
    Dwater = (5 Cwater)/Pi;
    Ewater = Dwater^3 + 2*Cwater;

    While[
      (Abs[point] < 100) && (iters++ < iterLim),

      ClassicalRungeKuttaCoefficients[4, prec_] :=
       With[{amat = {{1/2}, {0, 1/2}, {0, 0, 1}}, 
         bvec = {1/6, 1/3, 1/3, 1/6}, cvec = {1/2, 1/2, 1}}, 
        N[{amat, bvec, cvec}, prec]];

      f = 
       NDSolve[{SetPrecision[
          Derivative[1][y][x] == 
           Piecewise[{{(y[x] + x^3 + 3 z - 120*Ewater), 
              0 <= x <= 1}, {(y[x] + x^2 + 2 z), 
              1 <= x <= 2}, {(y[x] + x + z), 2 <= x <= 3}}], 40], 
         y[0] == 0}, y, {x, 0., 3.}, 
        Method -> {"ExplicitRungeKutta", "DifferenceOrder" -> 4, 
          "Coefficients" -> ClassicalRungeKuttaCoefficients}, 
        StartingStepSize -> 1/100];

      point = y[1] /. f[[1]];

      z++

      ]

     Print[point];



    prevTnew = Tnew;

    Tnew = 170 + 1.89*z;

    Cwater = 
     QuantityMagnitude[
      ThermodynamicData["Water", 
       "ThermalConductivity", {"Temperature" -> 
         Quantity[Tnew, "Kelvins"]}]];

    (*Note Tnew here!*)
    Print["Twater: ", Twater, " Tnew: ", Tnew];];

   (*End while*)

   ans = {"Tnew: ", Tnew, "Cwater: ", Cwater, "Dwater: ", Dwater, 
     "Ewater: ", Ewater};
   Print["The answer is:\n", ans];];

What I tried to do is to have "While Loop" inside a "While Loop" and instead of FindRoot, I use ReplaceAll, Here's the plan,

The inner loop is to find the value of "z" so that the "point" is 100, and the increment of "z" is 1 each time. After it reach the correct value of "z", the the command will continue to look for Tnew, and repeat again.

But the problem is

 point = y[1] /. f[[1]];

does not work properly in the While loop. I can do it outside of While loop but not inside. So, i'm stuck here

Hello Charles, thank you for your advice. You are correct about my background, that i have not been exposed to numerical analysis as much as I supposed to have. In the future, I will definitely take up PhD program if there is a chance.

I made the algorithm converge to an answer; I have no idea if it is correct. Here are the changes I made:

I replaced the statement Twater = Tnew, with Cwater = QuantityMagnitude[ ThermodynamicData["Water", "ThermalConductivity", {"Temperature" -> Quantity[Tnew, "Kelvins"]}]]; <-- note the Tnew here. I added the variable prevTnew, initialized it to 10^5, and set it to Tnew just before Tnew is computed. I changed the completion test to (Abs[prevTnew - Tnew] > tol) && (iters++ < iterLim). Also, please note I created the variable plotList, and initialized it to plotList = {}. Then after you compute point, I added AppendTo[plotList, point]. Later, you could put in Print[ListPlot[plotList]] if the reason you compute point is to plot it. I don't understand the 100 in point. Here is the new routine:

prog := Module[{Twater = 300., Tnew = 0, Cwater, Dwater, Ewater, 
    ClassicalRungeKuttaCoefficients, f, x, y, point, ans, tol = 0.001,
     iters = 0, iterLim = 100, prevTnew = 10^5, plotList = {}},
   $MinPrecison = 50; 
   Cwater = 
    QuantityMagnitude[
     ThermodynamicData["Water", 
      "ThermalConductivity", {"Temperature" -> 
        Quantity[Twater, "Kelvins"]}]];
   ans = "Twinkle, twinkle, little star,
         How I wonder what you are.
         Up above the world so high,
         Like a diamond in the sky.
         Twinkle, twinkle, little star,
         How I wonder what you are!";
   While[(Abs[prevTnew - Tnew] > tol) && (iters++ < iterLim), 
    Dwater = (5 Cwater)/Pi;
    Ewater = Dwater^3 + 2*Cwater;
    ClassicalRungeKuttaCoefficients[4, prec_] := 
     With[{amat = {{1/2}, {0, 1/2}, {0, 0, 1}}, 
       bvec = {1/6, 1/3, 1/3, 1/6}, cvec = {1/2, 1/2, 1}}, 
      N[{amat, bvec, cvec}, prec]];
    f = ParametricNDSolveValue[{SetPrecision[
        Derivative[1][y][x] == 
         Piecewise[{{(y[x] + x^3 + 3 z - 120*Ewater), 
            0 <= x <= 1}, {(y[x] + x^2 + 2 z), 
            1 <= x <= 2}, {(y[x] + x + z), 2 <= x <= 3}}], 40], 
       y[0] == 0}, y[3.], {x, 0., 3.}, z, 
      Method -> {"ExplicitRungeKutta", "DifferenceOrder" -> 4, 
        "Coefficients" -> ClassicalRungeKuttaCoefficients}, 
      StartingStepSize -> 1/10];
    point = {z /. FindRoot[f[z] == 100., {z, 1}, Evaluated -> False], 
      100.};
    AppendTo[plotList, point];
    Print[point];
    FindRoot[f[z] == 100., {z, 1}, Evaluated -> False];
    prevTnew = Tnew;
    Tnew = 
     170 + 1.89*z /. 
      FindRoot[f[z] == 100., {z, 1}, Evaluated -> False];
    Cwater = 
     QuantityMagnitude[
      ThermodynamicData["Water", 
       "ThermalConductivity", {"Temperature" -> 
         Quantity[Tnew, "Kelvins"]}]]; (* Note Tnew here! *)
    Print["Twater: ", Twater, " Tnew: ", Tnew];
    ];(*End while*)
   ans = {"Tnew: ", Tnew, "Cwater: ", Cwater, "Dwater: ", Dwater, 
     "Ewater: ", Ewater};
   Print["The answer is:\n", ans];
   ];

I was wondering if Module[] has something to do with it. It's just that the examples given in the documentation was just for simple tasks. I'm not in my Uni's lab right now (since the software was bought by Uni), but I'll get to it as soon as I can and let you know the result. Thank you very much.