b

$\frac{1}{F_a} = (\frac{\mu_g}{\mu_a} -1) (\frac{1}{R_1} - \frac{1}{R_2})$

$\frac{1}{F_l} = (\frac{\mu_g}{\mu_l} -1) (\frac{1}{R_1} - \frac{1}{R_2})$

$\frac{F_l}{F_a} = \frac{\frac{\mu_g}{\mu_a} -1}{\frac{\mu_g}{\mu_l} -1}$

$\frac{F_l}{10cm} = \frac{\frac{1.5}{1} -1}{\frac{1.5}{3} -1} = - 10 cm $

$\text{ Since focal length is negative hence its a divergent lens}$