Correct option is (b) 100
The mass of the earth, \(M_e = \frac{gR_e^2}G\)
The mass of the moon, \(M_m = \frac{g'R_m^2}{G}\)
\(\therefore \frac{M_e}{M_m} = \frac{gR_e^2}{g'R_m^2}\)
\(= \frac{g}{\frac g6}\times \frac{R^2}{\left(\frac R4\right)^2}\)
⇒ \(\frac{M_e}{M_m} = 96\approx100\)