The formula for the perimeter \( P \) of a square with side length \( s \) is given by:
\[ P = 4s \]
For this problem, the side length \( s \) is \( 3\sqrt{32} \).
First, we compute the perimeter:
\[ P = 4 \times (3\sqrt{32}) \]
Now, we can multiply the constants:
\[ P = 12 \sqrt{32} \]
Next, we need to simplify \( \sqrt{32} \):
\[ \sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2} \]
Now, substitute this back into the expression for the perimeter:
\[ P = 12 \times 4\sqrt{2} = 48\sqrt{2} \]
Thus, the perimeter of the square in simplest radical form is:
\[ \boxed{48\sqrt{2}} \]
The correct answer is option C: \( 48\sqrt{2} \).