Magnification $=-1$ i.e., image is inverted and of same size. It means, object is at $2\mathrm{F}$ and image is also at $2\mathrm{F}$.
Therefore, distance between $\mathit{u}$ and $\mathit{v}=4\mathrm{F}$ $=60\mathrm{cm}$.
$\therefore \mathit{F}=60∕4=15\mathrm{cm}$
Object will be at $=2\mathrm{F}=30\mathrm{cm}$.
If the object is shifted towards the lens by 20 $\mathrm{cm}$, the new object distance will be at $10\mathrm{cm}$ from the lens. This distance is less than the focal length, so the image formed in this case would be virtual, erect and on the same side.