b

$\text{Power of a lens } P = \frac{1}{f} = (\frac{\mu_g}{\mu_a} -1)(\frac{1}{R_1} - \frac{1}{R_2} = 5D$

$\text{when this is immerged in a liquid its focal length become -100 cm}$

$ P'= (\frac{\mu_g}{\mu_l} -1)(\frac{1}{R_1} - \frac{1}{R_2} = -1D$

$\text{Dividing these two equations }$

$\frac{P}{P'} = \frac{\frac{\mu_g}{\mu_a} -1}{\frac{\mu_g}{\mu_l} -1} = -5$

$\Rightarrow \mu = \frac{5}{3}$