The moment of inertia of the rod about $O$ is 

$\frac{1}{3}m{\ell ^2}$. The maximum angular speed of the rod is when the rod is instantaneously vertical. The energy of the rod in this condition is $\frac{1}{2}I{\omega ^2}\,$ where $I$ is the moment of inertia of the rod about $O.$ When the rod is in extreme portion, its angular velocity is zero momentarily. In this case, the energy of the rod is mgh where h is the maximum height to which the center of mass $(C.M)$ rises

$\begin{array}{l}
\therefore \,mgh = \frac{1}{2}I{\omega ^2} = \frac{1}{2}\left( {\frac{1}{3}m{l^2}} \right){\omega ^2}\\
 \Rightarrow h = \frac{{{\ell ^2}{\omega ^2}}}{{6g}}
\end{array}$