1981 AHSME Problems/Problem 9

Because the space diagonal of a cube with side length $s$ is $s \sqrt{3},$ the side length of the cube in this problem is $\frac{a}{\sqrt{3}}.$ The surface area of the cube is therefore $6(\frac{a}{\sqrt{3}})^2=6 \cdot \frac{a^2}{3}=\boxed{2a^2},$ which is answer choice $\boxed{\text{A}}.$