Answer
$\mu=\frac{1}{2} q \omega r^2 .$
Work Step by Step
The period of rotation is $T=2 \pi / \omega$, and this time all the charge passes any fixed point near the ring. The average current is $i=q / T=q \omega / 2 \pi$ and the magnitude of the magnetic dipole moment is
$$
\mu=i A=\frac{q \omega}{2 \pi} \pi r^2\\=\frac{1}{2} q \omega r^2 .
$$
