be. $Y.Y = \frac{{stress}}{{strain}} = \frac{F}{{A.\left( {\Delta L/L} \right)}},$

$F = applied\,force$

$A = \,area\,of\,cross - \,section\,of\,wire$

$Now,$

${Y_1} = {Y_2} \Rightarrow \frac{{FL}}{{\left( {{A_1}} \right)\left( {\Delta {L_1}} \right)}} = \frac{{FL}}{{\left( {{A_2}} \right)\left( {\Delta {L_2}} \right)}}$

Since load and length are same for both

$ \Rightarrow r_1^2\Delta {L_1} = r_2^2\Delta {L_2},$

$\left( {\frac{{\Delta {L_1}}}{{\Delta {L_2}}}} \right) = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2} = 4\,\Delta {L_1}\,:\,\Delta {L_2} = 4:1$