Explanation

(b) Since, voltage remains same in parallel, so,
$\ \ \ \ \ \ \ i \propto \frac{1}{R}$
$\Rightarrow \frac{i_{1}}{i_{2}} = \frac{R_{2}}{R_{1}}$
$\ \ \ \ \ \frac{i_{1}}{i_{2}} = \frac{\rho l_{2}/A_{2}}{\rho l_{1}/A_{1}}\ \ \ \ \ \ \ \ \ \ \ \ \ \left( \because R = \frac{\rho l}{A} \right)$
$\Rightarrow \frac{i_{1}}{i_{2}} = \frac{l_{2}}{l_{1}}\ \times \left( \frac{r_{1}}{r_{2}} \right)^{2}\ \ \ \ \ \left( \because A = \pi r^{2} \right)$
$\Rightarrow \frac{i_{1}}{i_{2}} = \frac{3}{4} \times \left( \frac{2}{3} \right)^{2}$
Hence, $\frac{i_{1}}{i_{2}} = \frac{1}{3}$