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 $${\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{C^2}/N - {m^2},{R_E} = 6.37 \times {10^6}m$$         ]

Solution :
Given,
Electric field $$E = 150\,N/C$$
Total surface charge carried by earth $$q = ?$$
According to Gauss’s law.
$$\eqalign{ & \phi = \frac{q}{{{ \in _0}}} = EA \cr & {\text{or,}}\,\,q = { \in _0}EA = { \in _0}E\pi {r^2} = 8.85 \times {10^{ - 12}} \times 150 \times {\left( {6.37 \times {{10}^6}} \right)^2}. \cr & \simeq 680\,KC \cr} $$
As electric field directed inward hence
$$q = - 680\,KC$$