The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about 150 N/C, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be: [Given: $$\epsilon_0 = 8.85 \times 10^{-12}$$ C$$^2$$/N-m$$^2$$, $$R_E = 6.37 \times 10^6$$ m]

The electric field is directed inward toward the center of the Earth, meaning it points in the direction opposite to the outward normal of the Earth's surface. By Gauss's law, the total charge enclosed is $$Q = \epsilon_0 \oint \vec{E} \cdot d\vec{A}$$. Since $$\vec{E}$$ is directed inward and $$d\vec{A}$$ points outward, $$\vec{E} \cdot d\vec{A} = -E\, dA$$, giving $$Q = -\epsilon_0 E \times 4\pi R_E^2$$.

Substituting the given values: $$Q = -8.85 \times 10^{-12} \times 150 \times 4\pi \times (6.37 \times 10^6)^2$$. Computing $$R_E^2 = (6.37)^2 \times 10^{12} = 40.58 \times 10^{12}$$ m$$^2$$, and $$4\pi R_E^2 = 4\pi \times 40.58 \times 10^{12} = 5.099 \times 10^{14}$$ m$$^2$$.

So $$Q = -8.85 \times 10^{-12} \times 150 \times 5.099 \times 10^{14} = -8.85 \times 150 \times 5.099 \times 10^{2} = -8.85 \times 76485 \approx -6.77 \times 10^5$$ C $$= -677$$ kC, which is approximately $$-680$$ kC.