Explanation

(d) Let $m_{1} = m$, then $m_{2} = 3m\ and\ m_{3} = 5m$. Again let $L_{3} = l$, then $L_{2} = 3l$ and $L_{1} = 5l$. If $\sigma$ be the density of copper, then
$A_{1} = \frac{m_{1}}{L_{1}\sigma} = \frac{m}{5l\sigma},A_{2} = \frac{3m}{3l\sigma}and\ A_{3} = \frac{5m}{l\sigma}$
Hence $A_{2} = \frac{5A_{1}}{3}and\ A_{3} = 25A_{1}$
$\therefore\ R_{1} = \frac{\rho L_{1}}{A_{1}} = \frac{\rho.5l}{A_{1}},R_{2} = \frac{\rho L_{2}}{A_{2}} = \frac{\rho.3l}{5A_{1}} = \frac{3}{25}R_{1},$
and $R_{3} = \frac{\rho L_{3}}{A_{3}} = \frac{\rho.l}{25A_{1}} = \frac{R_{1}}{125}$
$\therefore\ R_{1}:R_{2}:R_{3} = R_{1}:\frac{3}{25}R_{1}:\frac{R_{1}}{125} = 125:15:1$