Answer
See the detailed answer below.
Work Step by Step
First, we need to find the first image created by the lens.
We can use the thin lens formula
$$\dfrac{1}{f_1}=\dfrac{1}{s_1}+\dfrac{1}{s'_1}$$
Hence,
$$s_1'=\left[ \dfrac{1}{f_1}-\dfrac{1}{s_1} \right]^{-1}$$
Plug the known;
$$s_1'=\left[ \dfrac{1}{10}-\dfrac{1}{5} \right]^{-1}=\bf -10\;\rm cm$$
This means that the image is still on the left side of the lens at its focal point.
The height of this image is given by
$$m=\dfrac{-s_1'}{s_1}=\dfrac{h_1'}{h}$$
Hence,
$$h_1'=\dfrac{-s_1'h}{s_1}$$
Plug the known;
$$h_1'=\dfrac{-(-10)(1)}{5}=\bf 2\;\rm cm$$
It is an upright virtual image.
Now this image is the object of the mirror where $s_2=10+5=15$ cm, and $h_2=h_1'$.
So by the same approach,
$$s_2'=\left[ \dfrac{1}{f_2}-\dfrac{1}{s_2} \right]^{-1}$$
Plug the known;
$$s_2'=\left[ \dfrac{1}{-30}-\dfrac{1}{15} \right]^{-1}=\bf-10\;\rm cm$$
This means that the image is behind the mirror.
The height of this image is given by
$$h_2'=\dfrac{-s_2'h_1'}{s_s}=\dfrac{-(-10)(2)}{15}=\bf \frac{4}{3}\;\rm cm$$
This means it is an upright virtual image too.
This image is now the object for the lens where $s_3=5+10=15\;\rm cm$, so
$$s_3'=\left[ \dfrac{1}{f_1}-\dfrac{1}{s_3} \right]^{-1}$$
$$s_3'=\left[ \dfrac{1}{10}-\dfrac{1}{15} \right]^{-1}=\bf 30\;\rm cm$$
This means that this final image is 30 cm to the left from the lens.
And its height is then given by
$$h_3'=\dfrac{-s_3'h_2'}{s_3}=\dfrac{-(30)(\frac{4}{3})}{15}$$
$$h_3'=\bf -2.67\;\rm cm $$
Thus, the final image is inverted and its height is 2.67 cm. It is at a distance of 30 cm to the left from the lens.
