Focal length of plano-convex lens be ${\mathit{f}}_1$
$\frac{1}{{\mathit{f}}_1}=\left(1.5-1\right)\left(\frac{1}{15}-\frac{1}{\mathit{\infty }}\right)$
$\mathrm{Or},\frac{1}{{\mathit{f}}_1}=\frac{1}{30}$
$\therefore \mathrm{}{\mathit{f}}_1=30\mathrm{cm}$
Focal length of plano-concave lens be ${\mathit{f}}_2$
$\frac{1}{{\mathit{f}}_2}=\left(1.5-1\right)\left(\frac{1}{\mathit{\infty }}-\frac{1}{15}\right)$
$\therefore {\mathit{f}}_2=-30\mathrm{cm}$
since, parallel rays are incident on the plano convex lens it will form an image at a distance $30\mathrm{cm}$ from the lens (since, focal length is $30\mathrm{cm}$ ).
This image will act as an object for the plano concave lens.
Object distance $=30-20=10\mathrm{cm}$
Applying lens formula,
$\frac{1}{{\mathit{v}}_2}-\frac{1}{{\mathit{u}}_2}=\frac{1}{{\mathit{f}}_2}$

So, ${\mathit{v}}_2=15\mathrm{cm}$