Answer
See the detailed answer below.
Work Step by Step
The author asks for an image on the wall, which means he needs a real image on the wall, so we have to use a converging lens.
For lenses,
$$\dfrac{1}{f}=\dfrac{1}{s}+\dfrac{1}{s'}\tag 1$$
The author told us to put the lens between the object and the wall which means that
$$s+s'=2\;\rm m=200\;\rm cm\tag 2$$
Hence,
$$s'=200-s$$
Plugging into (1),
$$\dfrac{1}{f}=\dfrac{1}{s}+\dfrac{1}{200-s}$$
Plugging the known;
$$\dfrac{1}{32}=\dfrac{1}{s}+\dfrac{1}{200-s}=\dfrac{200-s+s}{(200-s)s}=\dfrac{200}{200s-s^2}$$
Thus,
$$ 200s-s^2 =(200)(32)=6400$$
$$s^2-200s+6400=0 $$
Hence,
$$s_1=\color{red}{\bf 40}{\;\rm cm}, {\rm and}\;\; \;\;s_2=\color{red}{\bf 160}{\;\rm cm}$$
Therefore, from (2),
$$s'_1=\color{red}{\bf 160}{\;\rm cm}, {\rm and}\;\; \;\;s'_2=\color{red}{\bf 40}{\;\rm cm}$$
Recalling that $m=-s'/s=h'/h$
So,
$$m_1=\color{blue}{\bf- 4} , {\rm and}\;\; \;\;m_2=\color{blue}{\bf- \frac{1}{4}}$$
Where $h'=mh$ whereas $h=2$ cm.
So that
$$h'_1=\color{red}{\bf 8}{\;\rm cm}, {\rm and}\;\; \;\;h'_2=\color{red}{\bf 0.5}{\;\rm cm}$$
And the two images are inverted.
