The normalized eigenfunctions for a rigid rotator that is free to rotate about its center of mass in three dimensions are the spherical harmonics $Y_l^m$ . (See Griffiths QM Problem 4.24). The given wave function can be expanded in terms of the spherical harmonics as follows

$\begin{align}\Psi(\theta,\phi) = \frac{1}{\sqrt{2}}\left(-i Y_1^{+1} + i Y_1^{-1}\right)\end{align}$
Therefore, measurement of $L_z$ will yield m = +1 or m = -1, which corresponds to $\hbar$ or $-\hbar$ . Therefore, answer (C) is correct.

Solution to 1996 Problem 52