The initial kinetic energy of the ball is $Mgh$ . The final kinetic energy is
$\begin{align*}\frac{1}{2} I \omega^2 + \frac{1}{2}M v^2\end{align*}$
$v = \omega R$ and for a hoop $I = M R^2$ , so the final kinetic energy is
$\begin{align*}\frac{1}{2} M R^2 \omega^2 + \frac{1}{2}M \left(\omega R\right)^2 = M R^2 \omega^2\end{align*}$
So,
$\begin{align*}\omega = \sqrt{g h /R^2}\end{align*}$
The angular momentum about the center of mass is
$\begin{align*}L = I \omega = \boxed{M R \sqrt{g h}}\end{align*}$
Therefore, answer (A) is correct.

Solution to 1986 Problem 76