Given, $\mathrm{f}_{\mathrm{a}}=0.15 \mathrm{~m}, \mu_{\mathrm{g}}=\frac{3}{2}, \mu_{\mathrm{w}}=\frac{4}{3}$
According to Lens maker's formula
$\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right) \quad$ where $\mu=\frac{\mu_{\mathrm{L}}}{\mu_{\mathrm{M}}}$
$\frac{1}{\mathrm{f}_{\mathrm{a}}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)$
$=\left(\frac{(3 / 2)}{1}-1\right) C$ where $C=\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
or $\quad \frac{1}{f_{a}}=\frac{C}{2} \quad \text{...(i)}$
Also, $\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)=\left(\frac{(3 / 2)}{(4 / 3)}-1\right) \mathrm{C}$
or $\quad \frac{1}{f_{\mathrm{W}}}=\frac{C}{8} \quad \text{...(ii)}$
From Eqs. (i) and (ii), we get
$$
\frac{\mathrm{f}_{\mathrm{w}}}{\mathrm{f}_{\mathrm{a}}}=\frac{\mathrm{C}}{2} \times \frac{8}{\mathrm{C}}=4
$$
or
$$
\begin{aligned}
\mathrm{f}_{\mathrm{w}} &=4 \mathrm{f}_{\mathrm{a}} \\
&=4 \times 0.15=0.6 \mathrm{~m}
\end{aligned}
$$