To solve this problem, we need to understand the concept of escape velocity.$\\$The escape velocity $v_e$ from the surface of a planet is given by the formula:$\\ v_e = \sqrt{\frac{2GM}{R}} \\$where $G$ is the gravitational constant, $M$ is the mass of the planet,$\\$and $R$ is the radius of the planet.$\\$For Earth, the escape velocity is $v = \sqrt{\frac{2GM_E}{R_E}}. \\$Given that the other planet has a radius four times that of Earth, $\\ R_{planet} = 4R_E. \\$The problem states that the mass density of the planet is the same as Earth's.$\\$Mass density $\rho$ is given by $\rho = \frac{M}{V},$ where $V$ is the volume.$\\$For a sphere, $V = \frac{4}{3}\pi R^3. \\$Since the density is the same, we have:$\\ \frac{M_{planet}}{V_{planet}} = \frac{M_E}{V_E}. \\$Substituting the volumes, we get:$\\ \frac{M_{planet}}{\frac{4}{3}\pi (4R_E)^3} = \frac{M_E}{\frac{4}{3}\pi R_E^3}. \\$Simplifying, we find:$\\ \frac{M_{planet}}{64R_E^3} = \frac{M_E}{R_E^3}. \\ M_{planet} = 64M_E. \\$Now, substitute $M_{planet}$ and $R_{planet}$ into the escape velocity formula:$\\ v_{e, planet} = \sqrt{\frac{2G(64M_E)}{4R_E}}. \\$Simplifying, we get:$\\ v_{e, planet} = \sqrt{\frac{128GM_E}{4R_E}}. \\ v_{e, planet} = \sqrt{32 \cdot \frac{2GM_E}{R_E}}. \\ v_{e, planet} = \sqrt{32} \cdot \sqrt{\frac{2GM_E}{R_E}}. \\ v_{e, planet} = \sqrt{32} \cdot v. \\ v_{e, planet} = 4v. \\$Therefore, the escape velocity from the surface of the other planet is $4v. \\$This corresponds to Option 4.