A 2 cm high object is placed 30 cm left of thin lens of focal length 20 cm as shown in the figure.

For lens 
${\mathrm{V}}_1=\frac{uf}{u-f}=\frac{30\times 20}{30-10}=60$ cm. (real image)
For mirror 
${\mathrm{V}}_2=\frac{10\times \left(-10\right)}{10-\left(-10\right)}=-5$ cm. (virtual image)
For lens 
${\mathrm{V}}_3=\frac{75\times 20}{75-20}=\frac{1500}{55}=22.27$ cm (real image)
$\mathrm{m}={\mathrm{m}}_1\times {\mathrm{m}}_2\times {\mathrm{m}}_3=2\times \frac{1}{2}\times \frac{4}{11}=\frac{4}{11}$
Image size = $\frac{4}{11}\times 2\mathrm{cm}=\frac{8}{11}\mathrm{cm}$
Final image is correct.