Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
As seen below, we can use any software calculator to get the final image object.
$$\color{blue}{\bf [b]}$$
Now we need to test our sketch. So we need to find the final image position.
To find the final image position, we need to find the position of the first image created by the converging lens by using the thin lens law.
$$\dfrac{1}{f_1}=\dfrac{1}{s_1}+\dfrac{1}{s_1'}$$
Thus,
$$s_1'=\left[\dfrac{1}{f_1}-\dfrac{1}{s_1}\right]^{-1}$$
Plugging the known;
$$s_1'=\left[\dfrac{1}{10}-\dfrac{1}{20}\right]^{-1}=\bf 20\;\rm cm$$
which is exactly as seen in the figure below.
The height of the first image is given by
$$m=\dfrac{h_1'}{h_1}=\dfrac{-s_1'}{s_1}$$
Hence,
$$h_1'=\dfrac{-h_1s_1'}{s_1}$$
Plugging the known;
$$h_1'=\dfrac{-(2)(20)}{(20)}=\bf-2\;\rm cm$$
This first image is the object of the second lens.
This image is at a distance of 40 cm from the second lens, so we need to use the lens thin law again to find the final position of the image.
$$s_2'=\left[\dfrac{1}{f_2}-\dfrac{1}{s_2}\right]^{-1}$$
Plugging the known;
$$s_2'=\left[\dfrac{1}{15}-\dfrac{1}{10}\right]^{-1}=\bf -30\;\rm cm$$
which is exactly, as seen in the figure below, on the first lens.
The height of the second image is given by
$$h_2'=\dfrac{-h_1's_2'}{s_2}=\dfrac{-(-2)(-30)}{(10)}=\bf -6\;\rm cm$$
And as we see, the two heights of the two images are matching the two images that are on the graph.
