Explanation

(d) Let $d$ be the density of the material of copper wire. Let $l_{1},l_{2}$ be the lengths of copper wires of diameter 1 mm and 2 mm respectively. As
Mass=volume$\times density =$($\pi D^{2}/4)l.d$
So, $1 = \frac{\pi\left( 10^{- 3} \right)^{2}}{4}l_{1} \times d = \pi \times \frac{\left( 2 \times 10^{- 3} \right)^{2}}{4}l_{2} \times d$
or $l_{1} = 4l_{2}$
Now, $R = \frac{\rho l}{\pi D^{2}/4}ie,R \propto \frac{l}{D^{2}}$
$\therefore\ \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} \times \frac{D_{2}^{2}\ \ }{D_{1}^{2}} = 4 \times 2^{2} = 16$