As we starch the wire, its length will increase but its radius will decrease keeping the volume constant

$\mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{f}}$

$\pi^2 l=\pi \frac{\mathrm{r}^2}{4} l_{\mathrm{f}}$

$l_{\mathrm{f}}=4 l$

$\frac{\mathrm{R}_{\text {nelv }}}{\mathrm{R}_{\text {old }}}=\left(\frac{4 l}{\frac{\mathrm{r}^2}{4}}\right) \frac{\mathrm{r}^2}{l}=16$

$\mathrm{R}_{\text {neer }}=16 \mathrm{R}$

$\therefore  \mathrm{x}=16$