Explanation

(d) When the currents are in the same direction then magnetic field
$B = \frac{\mu_{0}}{4\pi} \times \frac{2}{d}\ \left\lbrack i_{1} - i_{2} \right\rbrack$
$6 \times 10^{- 6} = \ \frac{\mu_{0}}{4\pi} \times \frac{2}{d}\ \left\lbrack i_{1} - i_{2} \right\rbrack\ \ \ \ldots(i)$
When the currents are in the reversed direction then magnetic field
$\ \ B_{2} = \frac{\mu_{0}}{4\pi} \times \frac{2}{d}\ \left\lbrack i_{1} - ( - i_{2}) \right\rbrack\ $
$\ 6 \times 10^{- 6} = \ \frac{\mu_{0}}{4\pi} \times \frac{2}{d}\ \left\lbrack i_{1} - i_{2} \right\rbrack\ \ \ldots(i)$
When the currents are in the reversed direction then magnetic field
$\ B_{2} = \frac{\mu_{0}}{4\pi} \times \frac{2}{d}\ \left\lbrack i_{1} - ( - i_{2}) \right\rbrack\ $
or $\ \ \ \ \ B_{2} = \frac{\mu_{0}}{4\pi} \times \frac{2}{d}\ \left\lbrack i_{1} + i_{2} \right\rbrack\ $
or $\ \ 3 \times 10^{- 5} = \frac{\mu_{0}}{4\pi} \times \frac{2}{d}\left\lbrack i_{1} + i_{2} \right\rbrack\ \ \ \ldots(ii)$
Dividing Eq. (i) by Eq. (ii)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{i_{1} - i_{2}}{i_{1} + i_{2}} = \frac{6 \times 10^{- 6}}{3 \times 10^{- 5}}$
or $\ \ \ \ \ \ \ \ \ \ \ \frac{i_{1} - i_{2}}{i_{1} + i_{2}} = \frac{2}{10}$
or $\ \ 5i_{1} - 5i_{2} = i_{1} + i_{2}$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ 4i_{1} = 6i_{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{i_{1}}{i_{2}} = \frac{6}{4}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{i_{1}}{i_{2}} = \frac{3}{2}$