The space diagonal (interior diagonal) of a cube has length $6$.  Find the $\textit{surface area}$ of the cube.
1. Let \( s \) be the edge length of the cube. The space diagonal (interior diagonal) of a cube can be expressed in terms of the edge length \( s \) using the Pythagorean theorem in three dimensions. The space diagonal \( d \) of a cube with edge length \( s \) is given by:
   \[
   d = s\sqrt{3}
   \]
2. We are given that the space diagonal \( d \) is 6. Therefore, we can set up the equation:
   \[
   s\sqrt{3} = 6
   \]
3. To solve for \( s \), divide both sides of the equation by \( \sqrt{3} \):
   \[
   s = \frac{6}{\sqrt{3}}
   \]
4. Rationalize the denominator:
   \[
   s = \frac{6}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}
   \]
5. The surface area \( A \) of a cube is given by \( 6s^2 \). Substitute \( s = 2\sqrt{3} \) into the formula:
   \[
   A = 6s^2 = 6(2\sqrt{3})^2
   \]
6. Calculate \( (2\sqrt{3})^2 \):
   \[
   (2\sqrt{3})^2 = 2^2 \cdot (\sqrt{3})^2 = 4 \cdot 3 = 12
   \]
7. Substitute back into the surface area formula:
   \[
   A = 6 \times 12 = 72
   \]

The final answer is \(\boxed{72}\)