Solution:

Focal length in air is given by,

$$ \frac{1}{f_a}=(_a\mu_g-1)\bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg)$$

The focal length of lens immersed in water is given by

$$ \frac{1}{f_l}=(_l \mu_g-1)\bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg)$$

where $$ R_1, R_2$$ are radii of curvatures of the two surfaces oflens

and $$_1 \mu_g$$ is refractive index of glass with respect to liquid.

Also $$ _1 \mu_g = \frac{_a \mu_g}{_a \mu_l } $$

Given, $$ _a\, n _g = 1.5, f_a = 12\, cm, _a n_l = \frac{4}{3} $$

$$\therefore \frac{f_1}{f_a}=\frac{(_a\mu_g-1)}{(_l\mu_g-1)}$$

$$\frac{f_1}{12}=\frac{(1.5-1)}{\bigg(\frac{1.5}{4/3}-1\bigg)}=\frac{0.5 \times 4 }{0.5}$$

$$ \Rightarrow f_1 = 4 \times 12 $$

$$ = 48\, cm $$