Answer
$16\;\rm cm$, $80\;\rm cm$
Work Step by Step
We have here a 1cm-tall object which is 20 cm to the left from a diverging lens. This lens is to the left from another lens and we don't know its focal length. Finally, there is a screen to the right of this unknown lens.
Let's rearrange them from left to right:
$\textbf{Object $\rightarrow$ diverging lens $\rightarrow$ unknown lens $\rightarrow$ screen.}$
Now we can use the thin lens formula to find the position of the first image from the diverging lens.
$$\dfrac{1}{f_1}=\dfrac{1}{s_1}+\dfrac{1}{s_1'}$$
Thus,
$$s_1'=\left[\dfrac{1}{f_1}-\dfrac{1}{s_1}\right]^{-1}$$
$$s_1'=\left[\dfrac{1}{-20}-\dfrac{1}{20}\right]^{-1}=\bf-10 \;\rm cm$$
This means that the first image, which is the object of the unknown lens, is on the same side of the object and it is 10 cm to the left from the diverging lens.
Let's assume that the distance between the two lenses is $x$ where we know that the distance between the original object and the screen is 110 cm.
So this first image is at a distance of
$s_2=10+x\tag 1$ to the left from the unknown lens, and the final image is then at a distance of $s_2'=90-x\tag 2$
See the figure below.
Now we need to find the height of this image,
$$m_1=\dfrac{-s_1'}{s_1}=\dfrac{h_1'}{h_1}$$
So,
$$h_1'=\dfrac{-s_1'h_1}{s_1}=\dfrac{-(-10)(1)}{20}=\bf 0.5\;\rm cm$$
It is an upright image.
The height of the final image is 2 cm, so
$$m_2=\dfrac{-s_2'}{s_2}=\dfrac{h_2'}{h_2}$$
where $h_2=h_1'$,
$$ \dfrac{-s_2'}{s_2}=\dfrac{h_2'}{h_1'}=\dfrac{2}{0.5}=4$$
So that
$$|s_2'|=|-4s_2|=4s_2$$
we neglected the signs since we are now dealing with lengths.
Plug into (2);
$$ 4s_2=90-x$$
Plug from (1),
$$ 4(10+x)=90-x$$
$$ 40+4x=90-x$$
Thus,
$$x=\bf 10\;\rm cm$$
This means that the distance between the two lenses is 10 cm and hence $s_2=10+x=20$ cm, and
$$s_2'=90-x=\color{red}{\bf 80}\;\rm cm$$
where $s_2'$ is the distance between the unknown lens and the screen.
Now we can find the focal length of this lens by using the thin lens formula again.
$$ f_2=\left[\dfrac{1}{s_2}+\dfrac{1}{s_2'}\right]^{-1}$$
$$ f_2=\left[\dfrac{1}{20}+\dfrac{1}{80}\right]^{-1}=\color{red}{\bf 16}\;\rm cm$$
