Case-I

If final image is at least distance of clear vision

$\mathrm{M}.\mathrm{P}.=\frac{\mathrm{L}}{{\mathrm{f}}_0}(1+\frac{\mathrm{D}}{{\mathrm{f}}_{\mathrm{e}}});375=\frac{150}{\mathrm{d}}[1+\frac{25}{{\mathrm{f}}_{\mathrm{e}}}]$

$\frac{375}{30}=1+\frac{25}{{\mathrm{f}}_{\mathrm{e}}}$

$\frac{345}{30}=\frac{25}{{\mathrm{f}}_{\mathrm{e}}}$

${\mathrm{f}}_0=\frac{750}{345}=2.17\mathrm{cm};{\mathrm{f}}_{\mathrm{e}}\approx 22\mathrm{mm}$

Case-II

If final image is at infinity

$\mathrm{M}.\mathrm{P}.=\frac{\mathrm{L}}{{\mathrm{f}}_0}\left(\frac{\mathrm{D}}{{\mathrm{f}}_{\mathrm{e}}}\right)=375$

f_e = 22 mm