Answer
$M=-9\times$.
Work Step by Step
Apply equation 25–2a and the provided magnification of the eyepiece to calculate its focal length.
$$f_e=\frac{N}{M}=\frac{25cm}{5}=5cm$$
The distance between the two lenses is the sum of the 2 focal lengths.
We set the sum of the focal lengths equal to the length of the telescope to calculate the focal length of the objective.
$$\ell=50cm=f_o+f_e=f_o+5cm$$
$$f_o=45cm$$
We know both focal lengths, so we may calculate the maximum magnification.
$$M=-\frac{f_o}{f_e}$$
$$M=-\frac{45cm}{5cm}=-9\times$$
