Young's modulus of copper $\left(Y_1\right) = 1.1\times {10}^{11} \mathrm{N} {\mathrm{m}}^{-2}$

Young's modulus of steel $\left(Y_2\right) = 1.9\times {10}^{11} \mathrm{N} {\mathrm{m}}^{-2}$

Let $d_1$ and $d_2$ be the diameters of copper and steel wires respectively.

Since the tension is the same in each wire, therefore each wire has the same extension. Also, as each wire is of the same length, the strain in each of them is the same.

$Y=\frac{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}}{{\displaystyle \raisebox{1ex}{$∆l$}\!\left/ \!\raisebox{-1ex}{$l$}\right.}}$

$\frac{Y_1}{Y_2}={\left(\frac{d_2}{d_1}\right)}^2$

$\frac{d_1}{d_2}=\sqrt{\frac{Y_2}{Y_1}}=\sqrt{\frac{19}{11}}$