Magnification $=-1$ i.e., image is inverted and of same size. It means, object is at $2F$ and image is also at $2F$.Therefore, distance between $u$ and $v=4F$ $=60\,\text{cm}$. $\therefore\; F=\frac{60}{4}=15\,\text{cm}$Object will be at $=2F=30\,\text{cm}$.If the object is shifted towards the lens by 20 $\text{cm}$, the new object distance will be at $10\,\text{cm}$ 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.