Answer
a) $5.86\;\rm cm$
b) $6.0\;\rm cm$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are given that $s+s'=300$ cm, and that
$$|m|=\dfrac{h'}{h}=\dfrac{98}{2}=\bf 49$$
where we know that the projector is creating a real image of a real object so that the image will be inverted.
Hence,
$$m=-\dfrac{s'}{s}=-49$$
Hence,
$$\dfrac{300-s}{s}= 49$$
Thus,
$$49s=300-s$$
So
$$s=\bf 6\;\rm cm$$
And hence,
$$s'=\bf 249\;\rm cm$$
We know, for thin lenses, that
$$\dfrac{1}{s}+\dfrac{1}{s'}=\dfrac{1}{f}$$
Thus, the focal length needed is given by
$$f=\left[ \dfrac{1}{s}+\dfrac{1}{s'}\right]^{-1}$$
Plugging the known;
$$f=\left[ \dfrac{1}{6}+\dfrac{1}{249}\right]^{-1}$$
$$f= \color{red}{\bf 5.86}\;\rm cm$$
$$\color{blue}{\bf b]}$$
We should put the lens at 6 cm from the slide.
$$x=s=\color{red}{\bf 6.0}\;\rm cm$$
