Answer
$3.55\;\rm m$
Work Step by Step
We know that the length of the telescope is given by the sum of the two focal lengths of the two lenses.
$$L=f_{\rm obj}+f_{\rm eye}\tag 1$$
we are given the focal length of the eye lens, so we have to work on the objective one.
We know that the magnification of the telescope is given by
$$M=-\dfrac{f_{\rm obj}}{f_{\rm eye}}=\dfrac{\theta_{\rm eye}}{\theta_{\rm obj}}$$
Hence,
$$f_{\rm obj}= f_{\rm eye}\dfrac{\theta_{\rm eye}}{\theta_{\rm obj}}\tag 2$$
we ignored the negative sign since we are dealing with lengths.
We are given $\theta_{\rm eye}$ and we know that $\theta_{\rm obj}$, for small angle approximation, is given by
$$\theta_{\rm obj}=\dfrac{D_{\rm Mars}}{R}$$
where $D$ is the diameter of Mars, and $R$ is the distance between it and our Earth.
Plugging into (2),
$$f_{\rm obj}=\dfrac{\theta_{\rm eye} f_{\rm eye}R}{D_{\rm Mars}} $$
Plugging into (1),
$$L=\dfrac{\theta_{\rm eye} f_{\rm eye}R}{D_{\rm Mars}}+f_{\rm eye} $$
Plugging the known and remember to convert the given angle to rad and other units to SI-system units.
$$L=\dfrac{\left[\dfrac{0.5^\circ\cdot \pi}{180^\circ}\right](25\times 10^{-3})( 1.1 \times 10^8\times 10^3)}{(6800\times 10^3)}+(25\times 10^{-3})$$
$$L=\color{red}{\bf 3.55}\;\rm m$$