A student has a mirror in a box. She cannot see its surface, but she can move a small lamp in front…

Step 1. The image distance. $m = -\dfrac{v}{u} = 3$, so $v = -3u = -3 \times (-20\ \text{cm}) = +60\ \text{cm}$. A positive $v$ means a virtual image behind the mirror.

Step 2. The focal length. We want $f$, so $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{60\ \text{cm}} - \dfrac{1}{20\ \text{cm}} = -\dfrac{1}{30\ \text{cm}}$. So $f = -30\ \text{cm}$.

Step 3. The radius and the kind. $R = 2f = -60\ \text{cm}$. A negative focal length means a concave mirror. This agrees with the enlarged, upright image, which only a concave mirror gives for an object inside the focus.

Step 4. The same size, real. We need $m = -1$, which means $v = u$. Then $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{2}{u} = \dfrac{1}{f}$, so $u = 2f = -60\ \text{cm}$. The lamp must be at the centre of curvature.