Answer
$7.5\cdot10^{-4}$ N$\cdot$m
Work Step by Step
The torque on the square current loop can be represented as $\vec{\tau} = \vec{m} \times \vec{B}$ where $\vec{m} = I\vec{A}$ is the magnetic moment. The torque on the current loop is
$\vec{\tau} = \vec{m} \times \vec{B} = (I\vec{A}) \times \vec{B} = I\vec{A}\vec{B}\sin(\theta)$
$\vec{\tau} = (500\cdot10^{-3})(0.05)^2(1.2)\sin(30) = 7.5\cdot10^{-4}$ N$\cdot$m
