Explanation

(c) According to lens formula $\frac{1}{f} = (\mu - 1)\left\lbrack \frac{1}{R_{1}} - \frac{1}{R_{2}} \right\rbrack$
The lens is plano-convex $i.e.,\ R_{1} = R$ and $R_{2} = \infty$
Hence $\frac{1}{f} = \frac{\mu - 1}{R} \Rightarrow f = \frac{R}{\mu - 1}$
Speed of light in medium of lens $v = 2 \times 10^{8}\ m/s$
$\Rightarrow \mu = \frac{c}{v} = \frac{3 \times 10^{8}}{2 \times 10^{8}} = \frac{3}{2} = 1.5$
If $r$ is the radius and $y$ is the thickness of lens (at the centre), the radius of curvature $R$ of its curved surface in accordance with the figure is given by
$R^{2} = r^{2} + (R - y)^{2} \Rightarrow r^{2} + y^{2} - 2Ry = 0$
Neglecting $y^{2};$ we get $R = \frac{r^{2}}{2y} = \frac{(6/2)^{2}}{2 \times 0.3} = 15\ cm$
Hence $f = \frac{15}{1.5 - 1} = 30\ cm$