In the adjoining figure, $PQ$ is a diagonal of the cube. If $PQ$ has length $a$, then the surface area of the cube is

$\text{(A)}\ 2a^2 \qquad \text{(B)}\ 2\sqrt{2}a^2 \qquad \text{(C)}\ 2\sqrt{3}a^2 \qquad \text{(D)}\ 3\sqrt{3}a^2 \qquad \text{(E)}\ 6a^2$
1. Let the side length of the cube be \( s \).
2. The diagonal \( PQ \) of the cube can be expressed in terms of the side length \( s \). In a cube, the space diagonal (which connects two opposite vertices) is given by \( s\sqrt{3} \). Therefore, we have:
   \[
   PQ = s\sqrt{3}
   \]
3. Given that \( PQ = a \), we can set up the equation:
   \[
   s\sqrt{3} = a
   \]
4. Solving for \( s \), we get:
   \[
   s = \frac{a}{\sqrt{3}} = \frac{a\sqrt{3}}{3}
   \]
5. The surface area of a cube is given by \( 6s^2 \). Substituting \( s = \frac{a\sqrt{3}}{3} \) into this formula, we get:
   \[
   \text{Surface Area} = 6 \left( \frac{a\sqrt{3}}{3} \right)^2
   \]
6. Simplifying the expression inside the parentheses:
   \[
   \left( \frac{a\sqrt{3}}{3} \right)^2 = \frac{a^2 \cdot 3}{9} = \frac{3a^2}{9} = \frac{a^2}{3}
   \]
7. Therefore, the surface area of the cube is:
   \[
   6 \cdot \frac{a^2}{3} = 2a^2
   \]

The final answer is \(\boxed{2a^2}\)