Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We need the most powerful lens with the shortest focal length to be used as an objective lens.
We assumed that the lenses are simple magnifiers whereas $M=25\;{\rm cm}/f$, so
$$f_{\rm 1}=\dfrac{25}{M_{\rm 1}}=\dfrac{25}{2}=\bf 12.5\;\rm cm$$
$$f_{\rm 2}=\dfrac{25}{M_{\rm 2}}=\dfrac{25}{4}=\bf 6.25\;\rm cm$$
Hence, $f_2$ is the objective lens, the most powerful with the shortest focal length.
$$\color{blue}{\bf [b]}$$
We know that the microscope magnification is given by
$$M=m_{\rm obj}M_{\rm eye}=\dfrac{L}{f_{\rm obj}}\dfrac{25\;\rm cm}{f_{\rm eye}}$$
$$M= \dfrac{L}{f_{\rm obj}}\dfrac{25\;\rm cm}{f_{\rm eye}}$$
To find the length of the tube, we need to solve for $L$,
$$L= \dfrac{Mf_{\rm eye}f_{\rm obj}} {25\;\rm cm} $$
Plugging the known (from part [a] above),
$$L= \dfrac{(12)(12.5)(6.25)} {25}=\color{red}{\bf 37.5}\;\rm cm $$
