Answer
$f = \frac{1}{2\pi}~\sqrt{\frac{5g}{2R}}$
Work Step by Step
The moment of inertia is $I = \frac{2}{5}MR^2$. The distance $d$ from the center of mass to the pivot is $R$.
We can find an expression for the frequency of small-angle oscillations as:
$f = \frac{1}{2\pi}~\sqrt{\frac{Mgd}{I}}$
$f = \frac{1}{2\pi}~\sqrt{\frac{MgR}{\frac{2}{5}MR^2}}$
$f = \frac{1}{2\pi}~\sqrt{\frac{5g}{2R}}$
