1. Identify the energy conversion

For a rod swinging about a horizontal axis through one end, the maximum angular speed occurs at the lowest position.

At that instant, the rod has maximum rotational kinetic energy: $K = \frac{1}{2} I \omega^2$

For a thin uniform rod about one end, $I = \frac{1}{3}ml^2$

So, $K = \frac{1}{2}\cdot \frac{1}{3}ml^2\omega^2 = \frac{1}{6}ml^2\omega^2$

2. At the highest point

At the extreme position, the rod momentarily comes to rest, so all this kinetic energy is converted into gravitational potential energy of the center of mass.

If the center of mass rises by height $h$, then $\Delta U = mgh$

By conservation of mechanical energy, $mgh = \frac{1}{6}ml^2\omega^2$

3. Solve for $h$

Cancelling $m$, $gh = \frac{1}{6}l^2\omega^2$

Hence, $h = \frac{1}{6}\frac{l^2\omega^2}{g}$

4. Match with the options

This corresponds to: $\boxed{\frac{1}{6}\,\frac{l^2\omega^2}{g}}$

So the correct option is C.

Note: Option C in the statement is written as $\frac{1}{6}\,\frac{l^2\omega^2}{g}$, which is dimensionally correct for height. Options involving $l\omega/g$ are dimensionally incorrect for height.