Given : ${\mathit{R}}_1=2\mathit{\Omega }$;
Let ${\mathit{l}}_1=\mathit{l}$ and ${\mathit{a}}_1={}^{\prime }\mathit{a}$ '
$\because$ The wire is stretched three time, so the new length $\left({\mathit{l}}_2\right)=3\mathit{l}$.
Let new area of crossection be ${\mathit{a}}_2$
$\because {\mathit{l}}_1\times {\mathit{a}}_1={\mathit{l}}_2\times {\mathit{a}}_2$
(Volume of wire is constant)
$\therefore \mathit{l}\times \mathit{a}=3\mathit{l}\times {\mathit{a}}_2$
$\Rightarrow {\mathit{a}}_2=\frac{\mathit{l}\times \mathit{a}}{3\mathit{l}}=\frac{\mathit{a}}{3}$
We know, $\mathrm{R}=\frac{\mathrm{\rho }\mathit{l}}{\mathit{a}}$

$\Rightarrow \frac{{\mathrm{R}}_2}{2}=\frac{3\mathit{l}}{\mathit{l}}\times \frac{\mathit{a}}{\mathit{a}∕3}$
$\Rightarrow \frac{{\mathrm{R}}_2}{2}=3\times 3$
$\Rightarrow {\mathrm{R}}_2=18\mathit{\Omega }$