Thank you for reporting, we will resolve it shortly
Q.
If the refractive index from air to glass is $\frac{3}{2}$ and
that from air to water is $\frac{4}{3}$, then the ratio of focal lengths of a glass lens in water and in air is
$n _{ w }=\frac{4}{3} \,\,\, n _{ g }=\frac{3}{2}$
$f _{ a }\left( n _{ g }-1\right)= f _{ w }\left(\frac{ n _{ g }}{ n _{ w }}-1\right)$
$ \frac{ f _{ w }}{ f _{ a }}=\frac{ n _{ g }-1}{\frac{ n _{ g }}{ n _{ w }}-1}=\frac{\frac{3}{2}-1}{\frac{\frac{3}{2}}{\frac{4}{3}}-1} $
$=\frac{\frac{3-2}{2}}{\frac{9}{8}-1}=\frac{\frac{1}{2}}{\frac{9-8}{8}}=\frac{\frac{1}{2}}{\frac{1}{8}}=\frac{8}{2}$
$ \therefore \frac{ f _{ w }}{ f _{ a }}=\frac{4}{1} $