Answer
${\bf 0.283}\;\rm A\cdot m^2$
Work Step by Step
We know that the magnetic torque on a magnetic dipole is given by
$$\vec \tau=\vec\mu\times \vec B=\mu B\sin\theta$$
So, the magnitude of the magnetic dipole moment is
$$\mu=\dfrac{\tau}{B\sin\theta}$$
Plug the given;
$$\mu=\dfrac{(0.02)}{(0.1)\sin45^\circ} $$
$$\mu=\color{red}{\bf 0.283}\;\rm A\cdot m^2$$
