Answer
$$\omega=\sqrt{\frac{g}{R}}$$
Work Step by Step
Gravitational force acting on the object pushes it to oscillate and, as a result, produces a torque. This restoring torque has a formula $\tau=-mgl$.
The lever arm here $l=R\sin\theta$, so we have $\tau=-mgR\sin\theta$. When $\theta$ is small enough, $\sin\theta=\theta$, so $$\tau=-mgR\theta$$
If we take $k'=mgR$, then $\tau=-k'\theta$. This equation is analogous to the restoring force $F_x=-kx$ in simple harmonic motion. In simple harmonic motion, $$\omega=\sqrt{\frac{k}{m}}$$
Here, if we replace $k$ with $k'$, $m$ is replaced with moment of inertia $I$, as they are analogous to each other. In this exercise, all mass is in the object, so $I=mR^2$ $$\omega=\sqrt{\frac{k'}{I}}=\sqrt{\frac{mgR}{mR^2}}=\sqrt{\frac{g}{R}}$$
