Let object be at a distance $x \mathrm{cm}$ from the pole. Then, distance of image from the pole will be $=15-x$.

Given:

Magnification, $m=2=\frac{-v}{u}$

$\Rightarrow 2=\frac{-(15-x)}{-x}$

$\Rightarrow 2x=15-x$

$$\begin{gathered} \Rightarrow 3x=15 \\ \Rightarrow x=5 \mathrm{cm} \end{gathered}$$

Now,

Distance of the image from the pole $=15-x=15-5=10 \mathrm{cm}$

Therefore, $v=10 \mathrm{cm}$

From mirror formula:

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$=\frac{1}{10}+\frac{1}{(-5)}=\frac{1-2}{10}=\frac{-1}{10}$

$\Rightarrow f=-10 \mathrm{cm}$