Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Let's assume that the lenses are simple magnifiers with $M=\dfrac{25\;\rm cm}{f}$.
Hence,
$$f_{\rm obj}=\dfrac{25}{M_{\rm obj}}\tag 1$$
and,
$$f_{\rm eye}=\dfrac{25}{M_{\rm eye}}\tag 2$$
where we know that the magnification of a telescope is given by
$$M=-\dfrac{f_{\rm obj}}{f_{\rm eye}}$$
Plugging from (1) and (2),
$$M=-\dfrac{\dfrac{25}{M_{\rm obj}}}{\dfrac{25}{M_{\rm eye}}}$$
$$M=-\dfrac{M_{\rm eye}}{M_{\rm obj}}$$
Noting that to maximize the telescope magnification, we should have ${M_{\rm eye}}\gt {M_{\rm obj}}$
So according to the given;
$$M=-\dfrac{5}{2}=\color{red}{\bf 2.5}$$
We ignored the negative sign since the magnification is an absolute value (but you still can use it since it means that the magnified image is upside down).
$$\color{blue}{\bf [b]}$$
We answered that above in part (a) when we wrote that we should have ${M_{\rm eye}}\gt {M_{\rm obj}}$.
Hence, the lens that should be used for the object is the one that has $2.0\times$ magnification.
$$\color{blue}{\bf [c]}$$
The minimum length of the telescope is given by the sum of the two focal lengths of the two lenses.
$$L=f_{\rm obj}+f_{\rm eye}$$
Plugging from (1) and (2),
$$L=\dfrac{25}{M_{\rm obj}}+\dfrac{25}{M_{\rm eye}}$$
Plug the known;
$$L=\dfrac{25}{2}+\dfrac{25}{5}=\color{red}{\bf 17.5}\;\rm cm$$
