Explanation

(c) Consider the refraction of the first surface, $i.e.,$ refraction from rarer medium to denser medium
$\frac{\mu_{2} - \mu_{1}}{R} = \frac{\mu_{1}}{- u} + \frac{\mu_{2}}{v_{1}} \Rightarrow \frac{\left( \frac{3}{2} \right) - \left( \frac{4}{3} \right)}{R} = \frac{\frac{4}{3}}{\infty} + \frac{\frac{3}{2}}{v_{1}} \Rightarrow v_{1} = 9R$
Now consider the refraction at the second surface of the lens, $i.e.,$ refraction from denser medium to rarer medium
$\frac{1 - \frac{3}{2}}{- R} = - \frac{\frac{3}{2}}{9R} + \frac{1}{v_{2}} \Rightarrow v_{2} = \left( \frac{3}{2} \right)R$
The image will be formed at a distance of $\frac{3}{2}R$. This is equal to the focal length of the lens