The resultant force on the current loop PQRS due to a long current carrying conductor is

Force on upper and lower parts becomes zero.

Net force on left and right part is given by,

 $\mathrm{F}=\frac{{\mu }_0{\mathrm{i}}_1{\mathrm{i}}_2\mathrm{l}}{2\pi }\left(\frac{1}{{\mathrm{r}}_1}-\frac{1}{{\mathrm{r}}_2}\right)=\frac{{\mu }_{0}20\times 20\times 15}{2\pi \times 2}-\frac{{\mu }_{\mathrm{o}}20\times 20\times 15}{2\pi \times 12}=5\times {10}^{-4}N$