The beem incident on lens $\mathrm{L}$ is parallel to principal axis. Hence, the lens forms an image ${\mathit{I}}_1$ at its focus, i.e., at a distanec $\mathit{O}{\mathit{I}}_1=20\mathrm{cm}$ from the lens.

The image ${\mathit{I}}_1$ is formed in front of mirror and hence, acts as a real source for the mirror. The concave mirror forms the image ${\mathrm{I}}_2$, whose distance from the mirror can be calculated as,
$\frac{1}{\mathit{v}}+\frac{1}{\mathit{u}}=\frac{1}{\mathit{f}}$
Here: $\mathit{u}=-30\mathrm{cm}$, and $\mathit{f}=-10\mathrm{cm}$
$\Rightarrow \frac{1}{\mathit{v}}=\frac{1}{\mathit{f}}-\frac{1}{\mathit{u}}$
$\Rightarrow \frac{1}{\mathit{v}}=-\frac{1}{10}+\frac{1}{30}$
$\Rightarrow \frac{1}{\mathit{v}}=\frac{1-3}{30}=-\frac{2}{30}$
$\Rightarrow \mathit{v}=-15\mathrm{cm}$
Hence, the final image is formed at a distance of $15\mathrm{cm}$ from the concave mirror, as shown in the following figure.