From the lens formula, we have:


$\frac{1}{\mathit{v}}-\frac{1}{\mathit{u}}=\frac{1}{\mathit{f}}$
$\frac{1}{\mathit{v}}=\frac{1}{\mathit{f}}+\frac{1}{\mathit{u}}$
$=\frac{1}{20}+\frac{1}{-60}$
$=\frac{3-1}{60}=\frac{2}{60}=\frac{1}{30}$
$\mathit{v}=+30\mathrm{cm}$
The positive sign indicates that the image is formed at the right of the lens.


The image ${\mathrm{I}}_1$ is formed behind the mirror and acts as a virtual oject for the mirror. The convex mirror forms the image ${\mathrm{I}}_2$, whose distance from the mirror can be determined as:

$\frac{1}{\mathit{v}}+\frac{1}{\mathit{u}}=\frac{1}{\mathit{f}}$
Here, $\mathit{u}=15\mathrm{cm}$
and, $\mathit{f}=\frac{\mathit{R}}{2}=10\mathrm{cm}$
$\frac{1}{\mathit{v}}=\frac{1}{\mathit{f}}-\frac{1}{\mathit{u}}$
$\mathit{v}=30\mathrm{cm}$
Hence, the final virtual image is formed at a distance $30\mathrm{cm}$ from the convex mirror.