The position of final image of an object O as shown in figure is

For plane surface,${\mathrm{h}}_{\mathrm{app}}=\mathrm{\mu h}=\frac{3}{2}\times 10=15\mathrm{cm}$

This first image is at a distance (15 + 3) cm from the plane mirror. So, mirror will make its second image at a distance 18 cm below the mirror or 21 cm below the plane surface. 

Now further applying ${\mathrm{d}}_{\mathrm{app}}=\frac{\mathrm{d}}{\mathrm{\mu }}=\frac{21}{1.5}=14\mathrm{cm}$ below the plane surface.