Given,
angular deceleration $\propto \sqrt{\text { angular speed }}$
i.e., $\quad \frac{-d \omega}{d t} \propto \sqrt{\omega}$
or $\quad \frac{-d \omega}{d t}=k \sqrt{\omega}$, where, $k$ is constant
or
$-\int_{\omega_0}^\omega \frac{d \omega}{\sqrt{\omega}}=k \int_0^t d t$
$-[2 \sqrt{\omega}]_{\omega_0}^\omega=k[t]_0^t$
$\Rightarrow \quad-2\left[\sqrt{\omega}-\sqrt{\omega_0}\right]=k t$

$$
\begin{aligned}
& \sqrt{\omega}-\sqrt{\omega_0}=-\frac{k t}{2} \\
\Rightarrow \quad & \sqrt{\omega}=\sqrt{\omega_0}-\frac{k t}{2}
\end{aligned}
$$

When $\omega=0$, then the total time of rotation,
$$
\tau=t=\frac{2 \sqrt{\omega_0}}{k}
$$
$\therefore$ Mean angular speed,
$$
\begin{aligned}
& < \bar{\omega}>=\frac{\int \omega d t}{\int d t}=\frac{\int_0^{2 \sqrt{\omega_0} / k}\left(\omega_0+\frac{k^2 t^2}{4}-k t \sqrt{\omega_0}\right) d t}{2 \sqrt{\omega_0} / k} \\
& =\frac{\left[\omega_0 t+\frac{k^2 t^3}{12}-\frac{k}{2} \sqrt{\omega_0} t^2\right]_0^{2 \sqrt{\omega_0} / k}}{2 \sqrt{\omega_0} / k}
\end{aligned}
$$

After solving, we get $=\frac{\omega_0}{3}$