Focal length of the biconvex lens is $f$. Radii are equal, i.e. $R_1=R_2=R$.
$$
{ }_a \mu_g=\frac{3}{2}
$$
$$
{ }_a \mu_w=\frac{4}{3}
$$
According to lens Maker's formula,
$$
\therefore \quad \frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
$$
Here, $R_1=R$ and $R_2=-R$.
$$
\begin{aligned}
\therefore & & \frac{1}{f} & =\left(\frac{3}{2}-1\right)\left(\frac{1}{R}+\frac{1}{R}\right)=\frac{1}{2} \times \frac{2}{R}=\frac{1}{R} \\
& \Rightarrow & & f=R
\end{aligned}
$$
When the lens is dipped in the water, then
$$
\begin{aligned}
\frac{1}{f_w} & =\left(\frac{{ }_a \mu_g}{{ }_a \mu_w}-1\right)\left(\frac{1}{R}+\frac{1}{R}\right) \\
\frac{1}{f_w} & =\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right) \times\left(\frac{1}{R}+\frac{1}{R}\right) \\
& =\left(\frac{3}{2} \times \frac{3}{4}-1\right) \times \frac{2}{R}=\left(\frac{9}{8}-1\right) \times \frac{2}{R} \\
\Rightarrow \quad \frac{1}{f_w} & =\frac{1}{8} \times \frac{2}{R}=\frac{1}{4 R}=\frac{1}{4 f} \Rightarrow f_w=4 f
\end{aligned}
$$