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}]]}]

Sorry, I should have pointed out that this was a conditional expression Im[b^2] != 0 || Re[b^2] >= 0], and Mathematica calculated this from Integrate[(p^2 (p^2 + 1)^(3/2))/(b^2 + p^2)^4, {p, 0, Infinity}].

I am unsure how to do a loop in Mathematica to construct a table of the value of the the minimum expression, and the value of "b" where the minimum occurs, either using NMinimize or Minimize i.e. vary alpha from 0.01,0.02,...0.1,0.2,0.3....1.0,1.1...2.0, and vary [Mu] from 0.01,0.02,..0.1,0.2,0.3....1.0,1.1,....10. It's an uneven grid, so alpha values go from 0.01 in increments of 0.01, then from 0.9 to 2.0 in increments of 0.1 etc.

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}]

An aside is the problem that evaluating D(-((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), b], then equating the result to == 0, and solving it -- the calculation doesn't complete even after 15 minutes. Same is the case if numerical differentiation is performed and the result equated to 0.