Explanation

(c) The wire are in parallel and ratio of their resistances are $3\ :4\ :5,$ hence currents in wires are in the ratio
$\frac{1}{3}:\frac{1}{4}:\frac{1}{5}$
$i_{1} = \frac{k}{3}$
$i_{2} = \frac{k}{4}$
$i_{3} = \frac{k}{5}$
Force between top and middle wire $F_{1} = \frac{\mu_{0}}{4\pi}.\frac{2i_{1}i_{2}}{r_{1}}$
$= \frac{\mu_{0}}{4\pi} \times \frac{2\left( \frac{1}{3} \right)\left( \frac{1}{4} \right)k^{2}}{r_{1}}$. Force between bottom and middle wire $F_{2} = \frac{\mu_{0}}{4\pi} \times \frac{\left( \frac{1}{4} \right)\left( \frac{1}{5} \right)k^{2}}{r_{2}}$. As the forces are equal and opposite so $F_{1} = F_{2} \Rightarrow \frac{r_{1}}{r_{2}} = \frac{5}{3}$