Here, for steel wire, Area of cross-section, $A_{1}=3.0 \times 10^{-5} \mathrm{m}^{2}$Stretching, $\Delta l_{1}=\Delta l$ (say) Stretching force on steel, $F_{1}=F$ For copper wire, length of wire, $l_{2}=3.5 \mathrm{m}$Area of cross-section, $A_{2}=4.0 \times 10^{-5} \mathrm{m}^{2}$ Stretching, $\Delta l_{2}=\Delta l$ (given);Stretching force on copper, $F_{2}=F$Let $Y_{1}$ and $Y_{2}$ be the Young's modulus of steel and copper wire respectively.and $Y_{2}= \frac{ F_{2} \times A_{2} }{ \frac{\Delta l_{2}}{l_{2}} }$$= \frac{F_{2} \times l_{2}}{A_{2} \times \Delta l_{2}}$$= \frac{F \times 3.5}{4 \times 10^{5} \times \Delta l} \ldots (ii)$Dividing (i) by (ii), we get $= \frac{18.8}{10.5} = 1.79 = 1.8$$\Rightarrow Y_{1} : Y_{2} = 1.8 : 1$