We know R= Aρl,R∝r2l As we starch the wire, its length will increase but its radius will decrease keeping the volume constant $\begin{aligned}
& \mathrm{V}{\mathrm{i}}=\mathrm{V}{\mathrm{f}} \
& \pi \mathrm{r}^2 l=\pi \frac{\mathrm{r}^2}{4} l_{\mathrm{f}} \
& l_{\mathrm{f}}=4 l \
& \frac{\mathrm{R}{\text {new }}}{\mathrm{R}{\text {old }}}=\left(\frac{4 l}{\frac{\mathrm{r}^2}{4}}\right) \frac{\mathrm{r}^2}{l}=16 \
& \mathrm{R}_{\text {new }}=16 \mathrm{R} \
\therefore & \mathrm{x}=16
\end{aligned}$