Given, ${ }^a \mu_g=\frac{3}{2}, f_{\text {air }}=f$
${ }^a \mu_w=\frac{4}{3}$
Using lens Maker's formula, when lens is in air,
$\begin{aligned} & \frac{1}{f_{\text {air }}}=\left({ }^a \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \Rightarrow \quad \frac{1}{f} & =\left(\frac{3}{2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \Rightarrow \quad \frac{1}{f} & =\frac{1}{2}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\end{aligned}$
When lens is immersed in water, then
$\frac{1}{f_w}=\left({ }^w \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Dividing Eq. (i) by Eq. (ii), we get
$\begin{aligned} \quad \frac{f_w}{f} & =\frac{1}{2\left(\mu_g-1\right)} \\ \Rightarrow \quad f_w & =\frac{f}{2\left(\frac{\mu_g}{\mu_w}-1\right)}=\frac{f}{2\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right)}=\frac{f}{\frac{1}{4}}=4 f \\ \Rightarrow \quad f_w & =4 f\end{aligned}$
$\therefore$ Percentage change in focal length
$\begin{aligned} & =\frac{f_w-f}{f} \times 100 \\ & =\frac{4 f-f}{f} \times 100=300 \%\end{aligned}$