Answer
$0.28$ A $\cdot$m$^2$
Work Step by Step
The torque on the magnet 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
$\vec{\tau} = \vec{m} \times \vec{B} = \vec{m}\vec{B}\sin(\theta)$
$\vec{m} = \displaystyle \frac{\vec{\tau}}{\vec{B}\sin(\theta)}$
$\vec{m} = \displaystyle \frac{0.020}{0.10\sin(45)} \approx 0.28$ A $\cdot$m$^2$
