A uniform rod of mass M and length L is hinged at one of its ends....

When the rod moves from the horizontal to the vertical position, its center of mass falls by a distance $h = \dfrac{L}{2}$.

The loss in gravitational potential energy is:
$\Delta U = Mg\left(\dfrac{L}{2}\right)$

This loss in potential energy is converted into rotational kinetic energy. The moment of inertia of a uniform rod about an axis through its end is $I = \dfrac{ML^2}{3}$.

Gain in rotational kinetic energy:
$\Delta K = \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}\left(\dfrac{ML^2}{3}\right)\omega^2$

By conservation of mechanical energy:
$Mg\left(\dfrac{L}{2}\right) = \dfrac{1}{2}\left(\dfrac{ML^2}{3}\right)\omega^2$
$MgL = \dfrac{ML^2}{3}\omega^2$
$\omega^2 = \dfrac{3g}{L}$
$\omega = \sqrt{\dfrac{3g}{L}}$

Assuming the center of mass falls by $L$ incorrectly gives $\sqrt{\dfrac{6g}{L}}$, and using the moment of inertia about the center ($\dfrac{ML^2}{12}$) incorrectly gives $\sqrt{\dfrac{12g}{L}}$.

Answer: $\sqrt{\dfrac{3g}{L}}$