The charge on a conductor sits on its surface with uniform density $\sigma =Q/(4\pi R^2)$. Consider a thin ring on the sphere at polar angle $\theta$, of radius $r=R\sin \theta$ and width $Rd\theta$. The charge on this ring is $dq=\sigma (2\pi R\sin \theta )(Rd\theta )$ and it circulates with period $T=2\pi /\omega$, giving current

$dI=\frac{dq}{T}=\sigma R^2\omega \sin \theta d\theta$

The magnetic moment contribution is $dM=dI\cdot \pi r^2=\sigma \pi R^4\omega {\sin }^3\theta d\theta$. Integrating with ${\int }_0^{\pi }{\sin }^3\theta d\theta =4/3$:

$M=\sigma \pi R^4\omega \cdot \frac{4}{3}=\frac{Q}{4\pi R^2}\cdot \pi R^4\omega \cdot \frac{4}{3}=\frac{Q{R}^2\omega }{3}$

For a uniform solid sphere, $L=\frac{2}{5}M{R}^2\omega$. The required ratio is

$\frac{M}{L}=\frac{Q{R}^2\omega /3}{(2/5)M{R}^2\omega }=\frac{5Q}{6M}=\frac{5}{3}\cdot \frac{Q}{2M}$

$\therefore \alpha =\frac{5}{3}\approx 1.67$.