Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We can use any software calculator to get the final image object, as seen below.
$$\color{blue}{\bf [b]}$$
We need to find the position of the first image of the first lens of 40 cm focal length.
Using thin lens formula,
$$\dfrac{1}{s_1}+\dfrac{1}{s_1'}=\dfrac{1}{f_1}$$
Hence,
$$s_1'=\left[\dfrac{1}{f_1}-\dfrac{1}{s_1}\right]^{-1}$$
Plugging the known;
$$s_1'=\left[\dfrac{1}{40}-\dfrac{1}{15}\right]^{-1}=\bf -24\;\rm cm$$
which is exactly as seen in the figure below.
The height of the first image is given by
$$m=\dfrac{h_1'}{h}=\dfrac{-s_1'}{s_1}$$
Hence,
$$h_1'=\dfrac{-hs_1'}{s_1}=\dfrac{-(2)(-24)}{15}=\bf 3.2\;\rm cm$$
This image is the object of the second lens of 20 cm focal length.
This image is at a distance of $s_2=24+10=\bf 34$ cm from this lens
Hence,
$$s_2'=\left[\dfrac{1}{f_2}-\dfrac{1}{s_2}\right]^{-1}$$
Plugging the known;
$$s_2'=\left[\dfrac{1}{20}-\dfrac{1}{34}\right]^{-1}$$
$$s_2'=\color{red}{\bf 48.6}\;\rm cm$$
which is exactly as seen in the figure below as well.
The height of the second image is given by
And as we see the two heights of the two images are matching those that are on the graph.
