Explanation

(d) Force on $SR$ and $PQ$ are equal but opposite so their net will be zero.
Force between two parallel conductors carrying currents $I_{1}\ and\ \ I_{2}$
$F = \frac{\mu_{0}}{2\pi}\frac{I_{1}I_{2}l}{r}$
Where $r =$ distance between two parallel conductors
$F_{PS} = \frac{10^{- 7} \times 2 \times 20 \times 20 \times 15 \times 10^{- 2}}{2 \times 10^{- 2}}$
$\ \ \ \ \ \ \ = 6 \times 10^{- 4}\ N$
$F_{QR} = \frac{10^{- 7} \times 2 \times 20 \times 20 \times 15 \times 10^{- 2}}{12 \times 10^{- 2}}$
$\ \ \ \ \ \ \ \ = 1 \times 10^{- 4}\ N$
$F_{net} = F_{PS} - F_{QR}$
$= 6 \times 10^{- 4} - 1 \times 10^{- 4} = 5 \times 10^{- 4}\ N$