For a given wire, its resistance

$\mathrm{R}\propto \frac{l}{\mathrm{A}}$
$\therefore R_1\propto \frac{l_1}{r_1^2}$

The wire is stretched such that $l_2=3l_1$

But density is unchanged.

$\therefore$Density $=\frac{M}{V}=\frac{M}{4\pi r_1^2{l}_1}=\frac{M}{4\pi r_2^2{l}_2}$

$\Rightarrow {\mathrm{r}}_2^2=\frac{l_1{\mathrm{r}}_1^2}{l_2}=\frac{r_1^2}{3} \dots ..\left(\mathrm{i}\right) \left(\because l_2=3l_1\right)$

Now, ${\mathrm{R}}_2\propto \frac{l_2}{{\mathrm{r}}_2^2}$

$\therefore {\mathrm{R}}_2\propto \frac{9l_1}{{\mathrm{r}}_1^2} ....\left(\mathrm{ii}\right)$

From $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$

$\therefore {\mathrm{R}}_2:{\mathrm{R}}_1=9:1$