Answer
$$
\tau=2 \pi \sqrt{\frac{k_G^2+d^2}{g d}}
$$
Work Step by Step
$$
\begin{gathered}
↻+\Sigma M_O=I_O \alpha ; \quad-m g d \sin \theta=\left[m k_G^2+m d^2\right] \ddot{\theta} \\
\ddot{\theta}+\frac{g d}{k_G^2+d^2} \sin \theta=0
\end{gathered}
$$
However, for small rotation $\sin \theta \approx \theta$. Hence
$$
\ddot{\theta}+\frac{g d}{k_G^2+d^2} \theta=0
$$
From the above differential equation, $\omega_n=\sqrt{\frac{g d}{k_G^2+d^2}}$.
$$
\tau=\frac{2 \pi}{\omega_n}=\frac{2 \pi}{\sqrt{\frac{g d}{k_G^2+d^2}}}=2 \pi \sqrt{\frac{k_G^2+d^2}{g d}}
$$
