If the distance between object and its two times magnified virtual...

Let object be at a distance $xcm$ from the pole. Then, distance of image from the pole will be $= 15 - x$.

Given:

Magnification, $m = 2 = \dfrac{- v}{u}$

$⇒ 2 = \dfrac{- ( 15 - x )}{- x}$

$⇒ 2x = 15 - x$

$⇒ 3x = 15 ⇒ x = 5cm$

Now,

Distance of the image from the pole $= 15 - x = 15 - 5 = 10cm$

Therefore, $v = 10cm$

From mirror formula:

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

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

$⇒ f = - 10cm$