The eigenstates with $m = 3$ are $Y_l^3$ where $l \geq 3$ . Therefore, the desired probability is
$\begin{align*}\sum_{l = 3}^{\infty} \left|\int \left(Y_l^3\right)^* \psi(\theta, \phi)\right|^2\end{align*}$
Using the fact that the spherical harmonics are orthogonal and normalized, this reduces to
$\begin{align*}&\left|\int \left(Y_4^3\right)^* \psi(\theta, \phi)\right|^2 + \left|\int \left(Y_6^3\right)^* \psi(\theta,...$
Therefore, answer (E) is correct.

Solution to 1992 Problem 28