The Bohr energies for an electron orbiting a nucleus with charge $Ze$ and mass $m_n$ is
$\begin{align*}E_n = \frac{- \mu Z^2 e^2}{2 \hbar^2 (4 \pi \epsilon_0)^2 n^2}\end{align*}$
where $\mu$ is the reduced mass of the electron and the nucleus. In this case $n = 2$ , $\mu = m_e/2$ . Therefore, the energy is $1/8$ of the ground-state energy of hydrogen. Therefore, answer (E) is correct.

Solution to 1992 Problem 30