Given, ratio of radius of two wires $r_1: r_2=2: 1$
$\Rightarrow \quad \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{2}{1}$
and ratio in their lengths, $1_1: 1_2=1: 2$
$\Rightarrow \quad \frac{l_1}{1_2}=\frac{1}{2}$
When same force is applied on them, then ratio r change in their length $\frac{\Delta \mathrm{l}_1}{\Delta \mathrm{l}_2}=$ ?
We know that, Young's modulus
$\begin{aligned}
& \mathrm{Y}=\frac{\mathrm{Fl}}{\mathrm{A} \Delta \mathrm{l}} \Rightarrow \Delta \mathrm{l}=\frac{\mathrm{Fl}}{\mathrm{AY}} \\
& \Delta \mathrm{l} \propto \frac{1}{\mathrm{~A}} \\
\therefore \quad & \frac{\Delta \mathrm{I}_1}{\Delta \mathrm{I}_2}=\frac{\mathrm{l}_1}{\mathrm{~A}_1} / \frac{\mathrm{l}_2}{\mathrm{~A}_2}=\frac{\mathrm{l}_1 \mathrm{~A}_2}{\mathrm{l}_2 \mathrm{~A}_1} \\
\Rightarrow \quad & \frac{\Delta \mathrm{l}_1}{\Delta \mathrm{I}_2}=\frac{\mathrm{l}_1}{\mathrm{l}_2} \cdot \frac{\mathrm{r}_2^2}{\mathrm{r}_1^2}=\frac{1}{2} \cdot\left(\frac{1}{2}\right)^2=\frac{1}{8}
\end{aligned}$