There are actually not too many combinations of digits to test (14 to be exact), but  I had written a program anyway, so I thought I might as well show all the possible combinations along the way, as if done manually. In each case the result of the computation is shown and then shown again if the product of the smallest, the middle and the square of the largest, digits has the same digits as used in the calculation:

 1 3 5   75

 1 3 7   147

 1 3 9   243

 1 5 7   245

 1 5 9   405

 1 7 9   567

 2 4 6   288

 2 4 8   512

 2 6 8   768

 3 5 7   735

 735

 3 5 9   1215

 3 7 9   1701

 4 6 8   1536

 5 7 9   2835

The solution, 735 is shown, beneath its three digits in numeric sequence.

 For a = 1 To 5

   For b = a + 2 To 9 Step 2

    For c = b + 2 To 9 Step 2

      p = a * b * c * c

      Text1.Text = Text1.Text & Str(a) & Str(b) & Str(c)

      Text1.Text = Text1.Text & "  " & Str(p) & crlf

      ps$ = Str(p)

      If InStr(ps, LTrim(Str(a))) > 0 Then

      If InStr(ps, LTrim(Str(b))) > 0 Then

      If InStr(ps, LTrim(Str(c))) > 0 Then

        Text1.Text = Text1.Text & ps & crlf

      End If

      End If

      End If

      DoEvents

    Next

   Next

 Next

The program has a bug (irrelevant to the solution as it turns out). Can you find it?

In declaring a solution the program assures that the three digits are all found in the product but does not check to be sure the result is not a 4-digit number with one of those digits a repeat. It just happens that a spurious solution was not found; of course it would have been obvious if a spurious solution had been found.