The energies can be found from
$\begin{align*}-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} = E_n \psi\end{align*}$
Evaluating the second derivative, we find that
$\begin{align*}-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} = \frac{\hbar^2 n^2 \pi^2}{2 m a^2} \psi\end{align*}$
Therefore, answer (B) is the only one that is correct.

Solution to 1992 Problem 53