Answer
a) $14\;\rm cm$
b) $-1.667\;\rm cm$
Work Step by Step
$$\color{blue}{\bf [a]}$$
First, we need to understand this system.
We have an object that is $L$ distance to the left from the 7-cm focal length lens. And then another lens to the right from the first lens which has a focal length of 15 cm where the distance between the two lenses is 20 cm.
Now we need to write these data in formulas,
We know that
$s_1=L$, $s_1'=$unknown, $s_2=$unknown, $s_2'=-10$ cm [to the right from the second lens that has a 15-cm focal length].
Now we need to use the thin lens formula to find $s_2$,
$$\dfrac{1}{f_2}=\dfrac{1}{s_2}+\dfrac{1}{s_2'}$$
$$s_2=\left[ \dfrac{1}{f_2}-\dfrac{1}{s_2'} \right]^{-1}$$
$$s_2=\left[ \dfrac{1}{15}-\dfrac{1}{-10} \right]^{-1}=\bf 6 \;\rm cm$$
This means that the first image position is 6 cm to the left from the 15-cm focal length lens. Thus, the first image is 14 cm to the right from the 7-cm focal length lens.
So that
$$s_1'=20-6=\bf 14\;\rm cm$$
Now we can use the thin lens formula for the first lens to find $L$,
$$\dfrac{1}{f_1}=\dfrac{1}{s_1}+\dfrac{1}{s_1'}$$
where $s_1=L$,
$$\dfrac{1}{f_1}=\dfrac{1}{L}+\dfrac{1}{s_1'}$$
Hence,
$$L=\left[ \dfrac{1}{f_1}-\dfrac{1}{s_1'}\right]^{-1}$$
Plugging the known;
$$L=\left[ \dfrac{1}{7}-\dfrac{1}{14}\right]^{-1}=\color{red}{\bf 14}\;\rm cm$$
$$\color{blue}{\bf [b]}$$
To find the height of the final image, we need to use the formula of
$$m=m_1m_2=\dfrac{h'}{h}$$
where $m=-s'/s=h'/h$, thus
$$\dfrac{h'}{h}=\left[ \dfrac{s_1'}{s_1}\right]\left[\dfrac{s_2'}{s_2} \right]$$
$$h'=h\left[ \dfrac{s_1'}{s_1}\right]\left[\dfrac{s_2'}{s_2} \right]$$
Plugging the known;
$$h'=(1)\left[- \dfrac{14}{14}\right]\left[-\dfrac{-10}{6} \right]$$
$$h'=\color{red}{\bf -1.667}\;\rm cm$$
It is an inverted image.
