Length of the steel wire, $L_{1}=4.7 m$

Area of cross-section of the steel wire, $A_{1}=3.0 \times 10^{-5} m ^{2}$

Length of the copper wire, $L_{2}=3.5 m$

Area of cross-section of the copper wire, $A_{2}=4.0 \times 10^{-5} m ^{2}$

Change in length $=\Delta L_{1}=\Delta L_{2}=\Delta L$

Force applied in both the cases $=F$

Young's modulus of the steel wire:

$Y_{1}=\frac{F_{1}}{A_{1}} \times \frac{L_{1}}{\Delta L}$

$=\frac{F \times 4.7}{3.0 \times 10^{-5} \times \Delta L} \ldots(i)$

Young's modulus of the copper wire:

$Y_{2}=\frac{F_{2}}{A_{2}} \times \frac{L_{2}}{\Delta L_{2}}$

$=\frac{F \times 3.5}{4.0 \times 10^{-5} \times \Delta L}\dots (ii)$

Dividing ($i$) by ($ii$), we get:

$\frac{Y_{1}}{Y_{2}}=\frac{4.7 \times 4.0 \times 10^{-5}}{3.0 \times 10^{-5} \times 3.5}=1.79: 1$

The ratio of Young's modulus of steel to that of copper is $1.79: 1$