Escape velocity from the Earth's surface is $v_e = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2G(\frac{4}{3}\pi R^3 \rho)}{R}} = \sqrt{\frac{8}{3}G\pi R^2 \rho} = R \sqrt{\frac{8}{3}G\pi \rho}$. Since $v_e \propto R$ for constant density, if $R' = 4R$, then $v_e' = 4v_e$. Wait, checking the calculation: $v_e = \sqrt{\frac{2G}{R} \cdot \frac{4}{3}\pi R^3 \rho} = \sqrt{\frac{8}{3}G\pi \rho} R$. If $R' = 4R$, then $v_e' = 4v_e$. Re-evaluating: The question asks for escape velocity. $v_e = \sqrt{2gR}$. Since $g = \frac{4}{3}\pi G \rho R$, $v_e = \sqrt{2(\frac{4}{3}\pi G \rho R)R} = R \sqrt{\frac{8}{3}\pi G \rho}$. Thus $v_e \propto R$. If $R$ becomes 4 times, $v_e$ becomes 4 times. However, the provided answer is 2v. Let's re-check: $v_e = \sqrt{\frac{2GM}{R}}$. $M = \rho \cdot \frac{4}{3}\pi R^3$. $v_e = \sqrt{\frac{2G \rho \frac{4}{3}\pi R^3}{R}} = \sqrt{\frac{8}{3}G\pi \rho} R$. If $R$ is 4 times, $v_e$ is 4 times. Perhaps the question implies mass is constant? No, it says same density. The provided answer is 2v, which would imply $v_e \propto \sqrt{R}$? No, that's incorrect. Given the provided answer is 4v, but the text says 2v. Let's re-read: 'radius four times'. $v_e \propto R$. So 4v.