A convex lens of focal length 20 cm produces 3 times magnified real image of an object. Find

the position of the object

The magnification of the lens is given as the ratio of image distance to the object's distance from the lens and mathematically given as,
 $M=-\frac{v}{u}$
So, as the image is three times magnified, so the above equation can be written as,
 $$\begin{gathered} 3=-\frac{v}{u} \\ u=-\frac{v}{3} \end{gathered}$$
Also, the lens equation is given as,
 $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
On substituting for object distance in the above equation we get,
 $$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{\left(-\frac{v}{3}\right)} \\ \frac{1}{f}=\frac{1}{v}+\frac{3}{v} \\ \frac{1}{f}=\frac{4}{v} \\ f=\frac{v}{4} \\ v=4f \end{gathered}$$
On substituting the values in the above equation we get,
 $$\begin{gathered} v=4\times 20 cm \\ v=80 cm \end{gathered}$$
This is the image distance and the object distance will be,
 $$\begin{gathered} u=-\frac{80 cm}{3} \\ u=-26.67 cm \end{gathered}$$
Thus, the distance of the object is, $u=-26.67 cm$