If the results end up in a variable named t, then you might use

t = Flatten[t, 1];
t3 = Table[{t[[i, 1]], t[[i, 2]], b /. t[[i, 3]][[2]]}, {i, Length[t]}]
ListPointPlot3D[t3, AxesLabel -> {"\[Alpha]", "\[Mu]", "b"}, PlotRange -> All]
ListPlot3D[t3, PlotRange -> All]

with output

You might want to try the following using Do loops to see if things are working:

Do[Do[Print[{alpha, mu, 
    NMinimize[{(-((4*alpha*b^3)/(2 b + mu)^2) + (64 b^5 (6 b + 4 b^3 - 16 b^5 + (3 Sqrt[1/b^2] b \[Pi])/
               Sqrt[1 - b^2] - (6 ArcSin[b])/Sqrt[1 - b^2]))/((96 b^5 - 96 b^7) \[Pi]))/(1 + b^2), 
      0 <= b <= 0.9999999}, {b}]}],
      {alpha, N[Flatten[{Range[10]/100, Range[2, 20]/10}]]}],
       {mu, N[Flatten[{Range[10]/100, Range[2, 100]/10}]]}] 

But when you are comfortable that things are working, using a Table command is much better:

t = Table[{alpha, mu, 
   NMinimize[{(-((4*alpha*b^3)/(2 b + mu)^2) + (64 b^5 (6 b + 4 b^3 - 16 b^5 + (3 Sqrt[1/b^2] b \[Pi])/
              Sqrt[1 - b^2] - (6 ArcSin[b])/Sqrt[1 - b^2]))/((96 b^5 - 96 b^7) \[Pi]))/(1 + b^2), 
     0 <= b <= 0.9999999}, {b}]},
     {alpha, N[Flatten[{Range[10]/100, Range[2, 20]/10}]]},
     {mu, N[Flatten[{Range[10]/100, Range[2, 100]/10}]]}]

If you post the Mathematica code (using the Code Sample icon) for the NMinimize and D[...,b] code above, you're more likely to get responses. And is that (1/b^2)^(1/2)*b in the equation? That would simplify to 1 would it not? Having the code would eliminate such guesswork.