$r = \frac{{2R}}{L}x + R$
At equilibrium change in length of the wire
$\int\limits_0^l {dL = \int {\frac{{Mgdx}}{{\pi {{\left[ {\frac{{2R}}{L}x + R} \right]}^2}y}}} } $
Taking limit from $0$ to $L$
$\Delta L = \frac{{Mg}}{{\pi y}} - \frac{1}{{\left[ {\frac{{2Rx}}{L} + R} \right]_0^L}} \times \frac{L}{{2R}} = \frac{{MgL}}{{3\pi {R^2}y}}$
The equilibrium extended length of wire
$ = L + \Delta L$
$ = L + \frac{{MgL}}{{3\pi {R^2}Y}} = L\left( {1 + \frac{1}{3}\frac{{Mg}}{{\pi Y{R^2}}}} \right)$



[Young's modulus for copper and steel are $1 \times 10^{11} N / m ^2$ and $2 \times 10^{11} N / m ^2$ respectively]