2023:
The answer is $\frac{F}{ml}$.
The moment of inertia of the rod about the hinge is $\frac{ml^2}{3}$. The torque acting on the rod is $\tau = F \times l$. The angular acceleration of the rod is $\alpha = \frac{\tau}{I} = \frac{Fl}{ml^2/3} = \frac{3F}{ml}$.
2022:
The answer is $\sqrt{\frac{3g}{L}}$.
The moment of inertia of the rod about the hinge is $\frac{mL^2}{3}$. The angular momentum of the rod about the hinge is $L = I \omega = \frac{mL^2}{3} \omega$. The kinetic energy of the rod is $K = \frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{mL^2}{3} \omega^2 = \frac{mL^2 \omega^2}{6}$. The gravitational potential energy of the ball is $U = mgh = mgL$. The total energy
