To solve this problem, we need to find the temperature at which the root mean square (rms) speed of oxygen molecules is equal to the escape velocity from Earth.$\\$The escape velocity $v_{escape}$ from Earth is given by:$\\ v_{escape} = \sqrt{\frac{2GM}{R}} \\$where $G$ is the gravitational constant, $M$ is the mass of Earth, and $R$ is the radius of Earth.$\\$For Earth, $G = 6.674 \times 10^{-11}$ m$^3$ kg$^{-1}$ s$^{-2}, \\ M = 5.972 \times 10^{24}$ kg$, \\ R = 6.371 \times 10^6$ m$. \\$Substitute these values into the escape velocity formula:$\\ v_{escape} = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{6.371 \times 10^6}} \\ v_{escape} = \sqrt{2 \times 6.674 \times 5.972 \times 10^{13}} \\ v_{escape} = \sqrt{7.986 \times 10^7} \\ v_{escape} \approx 11186$ m/s$\\$The rms speed $v_{rms}$ of gas molecules is given by:$\\ v_{rms} = \sqrt{\frac{3k_B T}{m}} \\$where $k_B$ is Boltzmann's constant, $T$ is the temperature, and $m$ is the mass of one molecule.$\\$We set $v_{rms} = v_{escape}$ to find the temperature:$\\ \sqrt{\frac{3k_B T}{m}} = 11186 \\$Square both sides to eliminate the square root:$\\ \frac{3k_B T}{m} = 11186^2 \\$Solve for $T: \\ T = \frac{11186^2 \cdot m}{3k_B} \\$Substitute the given values:$\\ m = 2.76 \times 10^{-26}$ kg$, \\ k_B = 1.38 \times 10^{-23}$ J K$^{-1} \\ T = \frac{11186^2 \cdot 2.76 \times 10^{-26}}{3 \times 1.38 \times 10^{-23}} \\ T = \frac{1.251 \times 10^{9} \cdot 2.76 \times 10^{-26}}{4.14 \times 10^{-23}} \\ T = \frac{3.45476 \times 10^{-17}}{4.14 \times 10^{-23}} \\ T \approx 8.360 \times 10^4$ K$\\$Therefore, the temperature at which the rms speed of oxygen molecules is sufficient for escaping Earth's atmosphere is $8.360 \times 10^4$ K$. \\$This corresponds to Option 2.