$\Delta {L_s} = \frac{{{W_s}L}}{{{Y_s}A}}$                          $\left( {asY = \frac{{W/A}}{{\Delta L/L}}} \right)$

and that in the brass wire is   $\Delta {L_b} = \frac{{{W_b}L}}{{{Y_b}A}}$

But $\Delta {L_s} = \Delta {L_b}$                           $(given)$

$\therefore \frac{{{W_s}L}}{{{Y_s}A}} = \frac{{{W_b}L}}{{{Y_b}A}}\,\,or\,\,\frac{{{W_s}}}{{{W_b}}} = \frac{{{Y_s}}}{{{Y_b}}}$

$As\,\frac{{{Y_s}}}{{{Y_b}}} = 2$                            $(given)$

$\therefore \frac{{{W_s}}}{{{W_b}}} = \frac{2}{1}$