Answer
${\bf 7.5\times 10^{-4}}\;\rm N\cdot m$
Work Step by Step
We know that the torque exerted by a magnetic field on a current loop is given by
$$\vec \tau=\vec\mu\times \vec B=\mu B\sin\theta$$
Recalling that $\mu =IA$, so
$$\tau=IAB\sin\theta$$
Plug the given;
$$\tau=(500\times 10^{-3})(0.05^2)(1.2)\sin30^\circ $$
$$\tau=\color{red}{\bf 7.5\times 10^{-4}}\;\rm N\cdot m$$
