The harmonic oscillator states are nondegenerate. Therefore, we can apply nondegenerate to find the first-order correction to the $n = 2$ energy level due to the perturbation to the Hamiltonian.

$\begin{align*}E_n^1 &= \langle 2 | \Delta H | 2 \rangle \\&= \langle 2 | V\left(a + a^{\dagger} \right)^2 | 2 \rangle...$
Both $\langle2 | aa | 2 \rangle$ and $\langle 2 | a^{\dagger}a^{\dagger} | 2 \rangle$ must equal $0$ because of the orthognality of the eigenstates. Therefore,
$\begin{align*}E_n^1 &= V\langle 2 | aa^{\dagger} | 2 \rangle + V\langle 2 | a^{\dagger}a | 2 \rangle \\ &= V\sqrt{3}...$
Therefore, answer (E) is correct.

Solution to 2001 Problem 94