The energy levels of the hydrogen atom are
$\begin{align*}E_n = \frac{-\mu e^4}{2 \hbar^2 (4 \pi \epsilon_0)^2 n^2}\end{align*}$
So,
$\begin{align*}E_n = \frac{- A}{n^2}\end{align*}$
where
$\begin{align*}A = \frac{\mu e^4}{2 \hbar^2 (4 \pi \epsilon_0)^2}\end{align*}$
Therefore, answer (E) is correct.

Solution to 1986 Problem 19