Answer
$$
\omega_n=\sqrt{\frac{3 g\left(4 R^2-l^2\right)^{1 / 2}}{6 R^2-l^2}}
$$
Work Step by Step
Moment of inertia about point $O$ :
$$
\begin{gathered}
I_O=\frac{1}{12} m l^2+m\left(\sqrt{R^2-\frac{l^2}{4}}\right)^2=m\left(R^2-\frac{1}{6} l^2\right) \\
C+\Sigma M_O=I_O \alpha, \quad m g\left(\sqrt{R^2-\frac{l^2}{4}}\right) \theta=-m\left(R^2-\frac{1}{6} l^2\right) \ddot{\theta} \\
\ddot{\theta}+\frac{3 g\left(4 R^2-l^2\right)^{\frac{1}{2}}}{6 R^2-l^2} \theta=0
\end{gathered}
$$
From the above differential equation, $\omega_n=\sqrt{\frac{3 g\left(4 R^2-l^2\right)^{\frac{3}{2}}}{6 R^2-l^2}}$.
