Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
According to the right-hand rule, the forces exerted by the magnetic field are shown in the figure below. Assuming that each force is exerted on a small part of the loop of $ds$.
Hence, the net force exerted on the current loop is given by
$$\sum F_y=F_{\rm net}=\sum (F\sin\theta)$$
where $dF=BIds $; so $F=BI\int_0^{2\pi R} ds$
$$ F_{\rm net}=BI\sin\theta\int_0^{2\pi R} ds$$
Hence,
$$ \boxed{F_{\rm net}=2\pi BIR\sin\theta }$$
We can see that the $x$-components cancel each other while $y$-components are added to make the net force.
$$\color{blue}{\bf [b]}$$
Plug the known into the boxed formula above.
$$F_{\rm net}=2\pi (0.20)(0.5)(0.02)\sin20^\circ$$
$$F_{\rm net}=\color{red}{\bf 4.3}\;\rm mN$$
