Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We have here two cases,
$\bullet$ We can find the image formed by the lens alone.
$\bullet$ Or we can find the image formed by the mirror as an object for the lens and then find the final image formed by the lens.
Noting that the rays from the object will go in both ways, first directly through the lens, or directly toward the mirror and then reflected from the mirror toward the lens.
$$\textbf{The first case:}$$
We need to find the image formed by the lens by using the thin lens law.
$$\dfrac{1}{f_{lens}}=\dfrac{1}{s_{ lens}}+\dfrac{1}{s_{lens}'}$$
Thus,
$$s_{lens}'=\left[ \dfrac{1}{f_{lens}}-\dfrac{1}{s_{ lens}} \right]^{-1}$$
$$s_{lens}'=\left[ \dfrac{1}{10}-\dfrac{1}{5} \right]^{-1}=\bf -10\;\rm cm$$
This means that the image formed by the lens is on the same side of the object and it is 10 cm to the left from the lens. This means that the image is just on the mirror.
Now the object of the mirror is the image formed by the lens which is at zero distance from the mirror, $s_{mirro}=0$ cm, and hence $s'_{mirror}=0$ as well.
Therefore, in this case, the final image is 10 cm to the left from the lens.
The height of this image is given by
$$m=\dfrac{-s'}{s}=\dfrac{h'}{h}$$
Hence,
$$h'=\dfrac{-hs'}{s}=\dfrac{-(1)(-10)}{5}=\bf 2\;\rm cm$$
See the first figure below.
$$\textbf{The second case:}$$
We need to find the image formed by the mirror by using the thin lens law.
$$\dfrac{1}{f_{mirror}}=\dfrac{1}{s_{ mirror}}+\dfrac{1}{s_{mirror}'}$$
Thus,
$$s_{mirror}'=\left[ \dfrac{1}{f_{mirror}}-\dfrac{1}{s_{ mirror}} \right]^{-1}$$
$$s_{mirror}'=\left[ \dfrac{1}{10}-\dfrac{1}{5} \right]^{-1}=\bf -10\;\rm cm$$
This image is 10 cm to the left of the mirror which means it is 20 cm to the left from the lens.
Now we need to find its height.
$$h_1'=\dfrac{-hs'}{s}=\dfrac{-(1)(-10)}{5}=\bf 2\;\rm cm$$
where $1$ here refers to the first image.
Now this is the object of the lens.
Now we need to find the image formed by the lens, by the same approach s we did above,
$$s_{lens}'=\left[ \dfrac{1}{f_{lens}}-\dfrac{1}{s_{ lens}} \right]^{-1}$$
$$s_{lens}'=\left[ \dfrac{1}{10}-\dfrac{1}{20} \right]^{-1}=\bf 20\;\rm cm$$
Now the second image is 20 cm to the right from the lens and its height is then
$$h_2'=\dfrac{-h_1's_{lens}'}{s_{lens}}=\dfrac{-(2)(20)}{20}=\bf -2\;\rm cm$$
where $2$ here refers to the first image.
Therefore, the final image is 2 cm-height and it is inverted. Its position is 20 cm to the right from the lens.
See the second figure below.
$$\color{blue}{\bf [b]}$$
