Resolve T along the horizontal and the vertical directions. As the particle moves in a horizontal plane, net vertical fore on it is zero. Net horizontal force on it provides the necessary centripetal acceleration for circular motion. Apply Netwon's second law in the horizontal direction to get

$$T\sin \theta = m{\omega ^2}r = m{\omega ^2}(l\sin \theta )$$

$$T = m{\omega ^2}l$$

$$\omega = \sqrt {{T \over {ml}}} $$

Substituting the given values, we get

$$\omega = \sqrt {{{324} \over {0.5 \times 0.5}}} = 36$$ rad/s