For the given convex lens, focal length, $f=25 \mathrm{cm}$

Magnification of lens in terms of $u$ and $f$ is given by,

$m=\frac{f}{f+u}=\frac{25}{25+u} ...\left(\mathrm{i}\right)$

When the image is real, then $m=-2$ and $u=-x_1$

$\therefore$ From Eq. $\left(\mathrm{i}\right)$, $-2=\frac{25}{25-x_1}\Rightarrow x_1=\frac{75}{2}$

Again, when the image is virtual, then $m=2$ and $u=-x_2$

$\therefore$ From Eq. $\left(\mathrm{i}\right)$, $2=\frac{25}{25-x_2}\Rightarrow x_2=\frac{25}{2}$

Hence, the ratio of $x_2$ and $x_1$ is

$\therefore \frac{x_2}{x_1}=\frac{25/2}{75/2}=\frac{1}{3}$