Rays coming from object AB first refract from the lens and then reflect from the mirror.
Refraction from the lens
$\mu =-\text{20 cm,}\text{f}=+\text{15 cm}$
Using lens formula, $\frac{1}{\upsilon }-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\upsilon }-\frac{1}{-20}=\frac{1}{15}$
 $\upsilon =+\text{60 cm}$
and linear magnification, ${\text{m}}_1=\frac{\upsilon }{\text{u}}=\frac{+60}{-20}=-3$
i,e., first image formed by the lens will be at 60 cm from it ( or 30 cm from mirror ) towards left and 3 times magnified but inverted. Length of first image A₁ B₁ would be 1.2 x 3 = 3.6 cm ( inverted).
                    
Reflections from mirror Image formed by lens ( A₁ B₁) will behave like a virtual object for mirror at a distance of 30 cm from it as shown. Therefore u =+ 30 cm, f =- 30 cm.
Using mirror formula,$\frac{1}{\upsilon }+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
or, $\frac{1}{\upsilon }+\frac{1}{30}=-\frac{1}{30}$
 $\therefore \upsilon =-15\text{cm}$
and linear magnification,${\text{m}}_2=-\frac{\upsilon }{\text{u}}=-\frac{-15}{+30}=+\frac{1}{2}$
i.e., final image A' B' will be located at a distance of 15 cm from the mirror ( towards right) and since magnification is $+\frac{1}{2}$, length of final image would be $3·6\times \frac{1}{2}=1·8\text{cm.}$
 $\therefore {\text{A}}^{\prime }{\text{B}}^{\prime }=1·8\text{cm}$
Point B is 0.6 cm above the optic axis of mirror, therefore, its image B' would be $\left(0·6\right)\left(\frac{1}{2}\right)=0·3\text{cm.}$ above optic axis. Similarly, point A is 3 cm below the optic axis, therefore, its image A' will be $3\times \frac{1}{2}=1·5\text{cm.}$ below the optic axis as shown below

Total magnification of the image,
 $\text{m}={\text{m}}_1\times {\text{m}}_2=\left(-3\right)\left(+\frac{1}{2}\right)=-\frac{3}{2}$
 $\therefore {\text{A}}^{\prime }{\text{B}}^{\prime }=\left(\text{m}\right)\left(\text{AB}\right)=\left(-\frac{3}{2}\right)\left(1·2\right)=-1·8\text{cm}$
Note that, there is no need of drawing the ray diagram if not asked in the question. Note With reference to the pole an optical instrument ( whether it is a lens or a mirror) the coordinates of the object (X₀, Y₀) are generally known to us. The corresponding coordinates of image ( X_i, Y_i) are found as follows
 ${\text{X}}_1\text{is obtained using }\frac{1}{\upsilon }\pm \frac{1}{\text{u}}=\frac{1}{\text{f}}$
Here,$\upsilon$is actually X₀ and u is X ie, the above formula can
be written as $\frac{1}{{\text{X}}_{\text{i}}}\pm \frac{1}{{\text{X}}_0}=\frac{1}{\text{f}}$
Similarly, Y is obtained from $\text{m}=\frac{\text{I}}{0}$
Here, I is Y and O is Y i.e., the above formula can be written as $\text{m}={\text{Y}}_{\text{i}}/{\text{Y}}_0\text{or}{\text{Y}}_{\text{i}}=\text{m}{\text{Y}}_0$