Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
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
$$\tau=I_{\rm loop}A_{\rm loop}B\sin\theta$$
where $B$ is due to the wire, so $B=\dfrac{\mu_0 I_{wire}}{2\pi d}$,
$$\tau=I_{\rm loop}A_{\rm loop} \dfrac{\mu_0 I_{wire}}{2\pi d}\sin\theta$$
where $A=\pi r^2$;
$$\tau=\pi \;r^2_{\rm loop}I_{\rm loop} \dfrac{\mu_0 I_{wire}}{2\pi d}\sin\theta$$
$$\tau= r^2_{\rm loop}I_{\rm loop} \dfrac{\mu_0 I_{wire}}{2 d}\sin\theta$$
Plug the known;
$$\tau= (1\times 10^{-3})^2(0.20) \dfrac{(4\pi \times 10^{-7}) (2)}{2 (0.02)}\sin90^\circ$$
$$\tau=\color{red}{\bf 1.26\times 10^{-11}}\;\rm N\cdot m$$
$$\color{blue}{\bf [b]}$$
The loop will have no torque exerted on it when the angle between the wire's magnetic field and its axis is $\theta=0^\circ$ or $\theta=180^\circ$.
This occurs when the loop rotates $90^\circ$ clockwise or counterclockwise.
