Answer
The magnitude of the electric field just outside the Earth's surface would be $~~1.8\times 10^{-8}~V/m$
Work Step by Step
We can find the surface area of the Earth:
$Area = 4\pi~R^2$
$Area = (4\pi)~(6.37\times 10^6~m)^2$
$Area = 5.10\times 10^{14}~m^2$
We can find the total charge of this number of electrons:
$Q = (-1.6\times 10^{-19}~C)(5.10\times 10^{14})$
$Q = -8.16\times 10^{-5}~C$
We can find the magnitude of the electric field just outside the Earth's surface:
$E = \frac{1}{4\pi~\epsilon_0}~\frac{\vert Q \vert}{R^2}$
$E = \frac{1}{(4\pi)~(8.854\times 10^{-12}~C/V~m)} \cdot \frac{8.16\times 10^{-5}~C}{(6.37\times 10^6~m)^2}$
$E = 1.8\times 10^{-8}~V/m$
The magnitude of the electric field just outside the Earth's surface would be $~~1.8\times 10^{-8}~V/m$
