To solve this problem, we need to understand the formula for escape velocity.$\\$The escape velocity $v_e$ is given by:$\\ v_e = \sqrt{\frac{2GM}{R}} \\$where $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet.$\\$The mass $M$ can be expressed in terms of density $\rho$ and volume $V$ as:$\\ M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi R^3 \\$Substituting $M$ in the escape velocity formula, we get:$\\ v_e = \sqrt{\frac{2G \cdot \rho \cdot \frac{4}{3} \pi R^3}{R}} \\ v_e = \sqrt{\frac{8 \pi G \rho R^2}{3}} \\$Now, let's compare the escape velocities of Earth and the planet.$\\$For Earth:$\\ v_{e} = \sqrt{\frac{8 \pi G \rho_e R_e^2}{3}} \\$For the planet:$\\ v_{p} = \sqrt{\frac{8 \pi G \rho_p R_p^2}{3}} \\$Given that the radius and mean density of the planet are twice that of Earth:$\\ R_p = 2R_e \\ \rho_p = 2\rho_e \\$Substitute these into the planet's escape velocity formula:$\\ v_{p} = \sqrt{\frac{8 \pi G (2\rho_e) (2R_e)^2}{3}} \\ v_{p} = \sqrt{\frac{8 \pi G \cdot 2 \rho_e \cdot 4 R_e^2}{3}} \\ v_{p} = \sqrt{\frac{64 \pi G \rho_e R_e^2}{3}} \\ v_{p} = \sqrt{4 \cdot \frac{16 \pi G \rho_e R_e^2}{3}} \\ v_{p} = 2 \sqrt{\frac{16 \pi G \rho_e R_e^2}{3}} \\ v_{p} = 2 \sqrt{\frac{8 \pi G \rho_e R_e^2}{3}} \\ v_{p} = 2 v_{e} \\$Thus, the ratio of escape velocity at Earth to the escape velocity at the planet is:$\\ \frac{v_{e}}{v_{p}} = \frac{v_{e}}{2v_{e}} = \frac{1}{2} \\$Therefore, the correct option is $$Option 4: 1:2√2$.$