If the distance between object and its two times magnified virtual image produced by a curved mirror is $$15$$ cm, the focal length of the mirror must be:

$$|m| = 2$$ (Hence, mirror is concave)

$$\text{Distance between object and image} = 15\text{ cm}$$

For a virtual image in a mirror, the image is formed behind the mirror while the object is in front:

$$\text{Distance} = |v| + |u| = 15\text{ cm}$$

Magnification relation for mirrors: $$m = -\frac{v}{u} \implies |v| = |m||u| = 2|u|$$

$$2|u| + |u| = 15 \implies 3|u| = 15 \implies |u| = 5\text{ cm}$$

$$|v| = 2 \times 5 = 10\text{ cm}$$

$$u = -5\text{ cm},\quad v = +10\text{ cm}$$

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \implies \frac{1}{f} = \frac{1}{10} + \frac{1}{-5} = -\frac{1}{10} \implies f = -10\text{ cm}$$