Solution
= Solution
The energy fixes $l=2$, irrespective of the magnetic quantum number $m$. Therefore every prepared state is an eigenstate of $\hat L^2$ with
$$
L=\sqrt{l(l+1)}\hbar=\sqrt6\hbar.
$$
An immediate measurement of the total orbital-angular-momentum magnitude cannot yield $\hbar$, so the requested probability is
$$
\boxed{0}.
$$