(d) Mutual force between conductors $A\ and\ C$
$F_{1} = \frac{\mu_{0}}{2\pi}\frac{I_{1}I_{2}l}{r} = \frac{\mu_{0}}{2\pi}\frac{2 \times 3 \times 1}{0.05}$
$\ \ \ \ \ = 2.40 \times 10^{- 5}\ N\ (towards\ A)$
Mutual force between conductors $B\ and\ C$
$= \frac{\mu_{0}}{2\pi} \times \frac{4 \times 3 \times 1}{0.08}$
$\ \ \ \ \ = 3 \times 10^{- 5}\ N\ (towards\ B)$
Hence, the resultant force experienced by $C$
$= (3 - 2.4) \times 10^{- 5}\ N$
$= 0.6 \times 10^{- 5}\ N\ (towards\ B)$