A light rod of length l is rigidly attached to a small heavy block...

Let the mass of the heavy block be $m$. Since the system is released from rest in a horizontal position, it undergoes pure translational motion downwards under gravity.

When the hook has fallen a depth $h$, the block has also fallen a depth $h$. The velocity of the system just before the hook catches the ring is given by:
$v = \sqrt{2gh}$ (directed vertically downwards)

When the hook catches the smooth ring, it acts as a fixed pivot. At this instant, the rod is horizontal, and the velocity of the block is perpendicular to the rod. The impulsive force from the ring acts at the pivot, so it exerts zero torque about the pivot. Therefore, the angular momentum of the block about the ring is conserved.

Let $\omega$ be the angular velocity of the rod just after the catch. Conserving angular momentum about the ring:
$mvl = ml^2\omega \implies \omega = \dfrac{v}{l}$

The kinetic energy of the block just after the catch is:
$K = \dfrac{1}{2}m(\omega l)^2 = \dfrac{1}{2}mv^2 = mgh$

Since the block is attached to a light rigid rod, the condition for it to complete a vertical circle is that it must just reach the highest point of the circle. At the highest point, its velocity can be zero ($v_{top} \ge 0$).

The highest point of the circle is at a height $l$ above the horizontal level of the ring. The potential energy required to reach this highest point from the ring's level is $mgl$.

By conservation of mechanical energy:
$K \ge mgl$
$mgh \ge mgl$
$h \ge l$

Thus, the minimum value of $h$ required is $l$.

Answer: $l$