Answer
$L_0={I\over mh}$
Work Step by Step
The time period of a physical pendulum whose distance of centre of mass from the point of suspension is $h$ is \[T=2\pi \sqrt{I\over mgh}\]
If $L_0$ is the distance of the centre of oscillation from the point of suspension, the period can be computed using \[T=2\pi\sqrt{L_0\over g}\]
Equating the two expressions, we get \begin{align*}
2\pi\sqrt{I\over mgh}&=2\pi\sqrt{L_0\over g}\\
{I\over mgh}&={L_0\over g}\\
L_0&={I\over mh}
\end{align*}
