The gravitational pull of the moon is $\left(\frac{1}{6}\right)^{\text {th }}$ of the earth and mass of moon is $\left(\frac{1}{8}\right)^{\text {th }}$ of the earth. This implies that the

The acceleration due to gravity at a celestial body's surface is governed by $g = \frac{GM}{R^2}$, where $M$ is mass and $R$ is radius. Given that lunar gravity is one-sixth of Earth's, $g_m = \frac{1}{6} g_e$, and lunar mass is one-eighth of Earth's, $M_m = \frac{1}{8} M_e$.

Substituting into the gravity formula yields $\frac{GM_m}{R_m^2} = \frac{1}{6} \cdot \frac{GM_e}{R_e^2}$. After canceling $G$ and $M_e$ using the mass relation, $\frac{1}{8R_m^2} = \frac{1}{6R_e^2}$.

Cross-multiplying gives $6R_e^2 = 8R_m^2$, simplifying to $3R_e^2 = 4R_m^2$. Solving for the radii, $R_e = \sqrt{\frac{4}{3}} R_m$, confirming that Earth's radius is $\sqrt{4/3}$ times the Moon's.

Only option B matches this result.