The escape velocity of an object from the earth depends upon the mass of earth $(M)$, its mean density $(\rho)$, its radius $(R)$ and gravitational constant $(G)$, thus the formula for escape velocity is:

We know for escape velocity $\left(v_{e}\right)$ :-

$\begin{aligned} & \frac{G m M}{R^{2}}=\frac{1}{2} m v_{e}^{2} \\ \therefore v_{e} & =\sqrt{\frac{2 G M}{R}} \end{aligned}$

According to question:-

$M=\left(\frac{4}{3} \pi R^{3}\right) \cdot \rho$

$\longrightarrow$ Volume of earth

$\therefore M=\frac{4 \pi R^{3} P}{3}$

Substituting this in equation for $V_{e}:-$

$\begin{array}{l} V_{e}=\sqrt{\frac{2 G}{R} \times \frac{4 \pi R^{3} \rho}{3}} \\ \therefore V_{e}=R \sqrt{\frac{8 \pi}{3} G \rho} . \end{array}$

Hence, option $A$ is correct.