Given, $\mathit{A}=-40\mathrm{cm}$ and, $\mathit{f}=20\mathrm{cm}$
From the lens formula, we have:
$\frac{1}{\mathit{v}}-\frac{1}{\mathit{u}}=\frac{1}{\mathit{f}}$
$\Rightarrow \frac{1}{\mathit{v}}=\frac{1}{\mathit{f}}+\frac{1}{\mathit{u}}$
$\Rightarrow \frac{1}{\mathit{v}}=\frac{1}{20}+\frac{1}{\left(-40\right)}$
$\Rightarrow \frac{1}{\mathit{v}}=\frac{2-1}{40}=\frac{1}{40}$
$\mathit{v}=40\mathrm{cm}$
The positive sign describes that the image is formed to the right of the lens.

The image ${\mathit{I}}_1$ is formed behind the mirror and thus acts as a virtual source for the mirror. The convex mirror forms the image ${\mathrm{I}}_2$, whose distance from the mirror is given by:
$\frac{1}{\mathit{v}}+\frac{1}{\mathit{v}}=\frac{1}{\mathit{f}}$
Here: $\mathit{u}=25\mathrm{cm}$
$\mathit{f}=\frac{\mathit{R}}{2}=10\mathrm{cm}$
$\frac{1}{\mathit{v}}+\frac{1}{\mathit{v}}=\frac{1}{\mathit{f}}$
$\Rightarrow \frac{1}{\mathit{v}}=\frac{1}{\mathit{f}}-\frac{1}{\mathit{u}}$
$\Rightarrow \frac{1}{\mathit{v}}=\frac{1}{10}-\frac{1}{25}$
$\Rightarrow \frac{1}{\mathit{v}}=\frac{5-2}{50}=\frac{3}{50}$
$\Rightarrow \mathit{v}=+16.67\mathrm{cm}$
Hence, the final image is formed at a distance of $16.67\mathrm{cm}$ behind the convex mirror.