A nonconducting ring of radius r and mass m carries a total charge...

The rotating charge on the ring constitutes an equivalent electric current.

The equivalent current is given by $i = \dfrac{q}{T}$, where $T$ is the time period of rotation.

Since the angular speed is $\omega$, the time period is $T = \dfrac{2\pi}{\omega}$.

Therefore, the current is $i = \dfrac{q \omega}{2\pi}$.

The magnetic moment $\mu$ of a current loop is the product of the current and the area of the loop.

The area of the ring is $A = \pi r^2$.

Substituting the values, we get $\mu = i A = \left(\dfrac{q \omega}{2\pi}\right) (\pi r^2) = \dfrac{q \omega r^2}{2}$.