Microscope!

As the final image is formed at infinity, the image would have been on focus of eyepiece, that is at a distance of $5\text{ cm}$ from the eyepiece.

$\large \displaystyle \therefore \text{Image distance for the objective } = \text{Tube length(L)} - f_e\\ \large \displaystyle = 20 - 5 = 15\text{ cm}\\ \large \displaystyle \text{Using Cartesian sign convention for the objective lens, }\\ \large \displaystyle \frac{1}{u_0} = \frac{1}{v_0} - \frac{1}{f_0}\\ \large \displaystyle \frac{1}{u_0} = \frac{1}{15} - \frac{1}{1} = -\frac{14}{15}\\ \large \displaystyle \therefore u_0 = -\frac{15}{14} \text{ cm}.$

When the final image is formed at infinity, the magnifying power of the microscope will be,

$\large \displaystyle m = \frac{v_0}{u_0} \times \frac{D}{f_e}\\ \large \displaystyle = \frac{15}{\frac{15}{14}} \times \frac{25}{5}\\ \large \displaystyle m = \color{#D61F06}{\boxed{70}}.$