$T \cos \theta$ component will cancel $\mathrm{mg}$.
$T \sin \theta$ component will provide necessary centripetal force to the ball towards centre $C$.
$$
\begin{array}{rlrl}
& \therefore & T \sin \theta & =m r \omega^2=m(l \sin \theta) \omega^2 \\
& \text { or } & T & =m l \omega^2 \\
& \therefore & \omega & =\sqrt{\frac{T}{m l}} \\
& \text { or } & \omega_{\max } & =\sqrt{\frac{T_{\max }}{m l}}=\sqrt{\frac{324}{0.5 \times 0.5}} \\
& = & 36 \mathrm{rad} / \mathrm{s}
\end{array}
$$
$\therefore$ Correct option is (d).
Analysis of Question
(i) Question is simple.
(ii) This is called the conical pendulum.
(iii) The interesting fact in this problem is that $\omega$ or $T$ is independent of $\theta$.
$\omega \propto \sqrt{T}$
If $\omega$ is increased, $T$ will also increase.