For a lens,
Magnification $=\frac{\text { Size of image }}{\text { Size of object }}=\frac{v}{u}$
$\Rightarrow \quad m=\frac{I}{O}=\frac{v}{u}$
Here, $m= \pm 4$
For $m=-4,-4=\frac{v}{u} \Rightarrow v=-4 u$
Now by lens formula,
$\begin{array}{rlrl} & & \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \Rightarrow \frac{1}{-4 u}-\frac{1}{u}=\frac{1}{20} \quad(\because f=20 \mathrm{~cm}) \\ \Rightarrow \quad & \frac{-5}{4 u} & =\frac{1}{20} \Rightarrow u=\frac{-5 \times 20}{4}=-25 \mathrm{~cm}\end{array}$
So, object is at a distance of $25 \mathrm{~cm}$ from lens when image is real.
For, $m=+4,+4=\frac{v}{u} \Rightarrow v=4 u$
Again from lens formula,
$\begin{aligned} \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \\ \Rightarrow \quad \frac{1}{4 u}-\frac{1}{u} & =\frac{1}{20} ; \Rightarrow \frac{-3}{4 u}=\frac{1}{20} \\ \text { or } u=\frac{-3}{4} \times 20 & =-15 \mathrm{~cm}\end{aligned}$
So, object is at a distance of $15 \mathrm{~cm}$ from the lens, when image is virtual.