Supposing that the earth has a surface charge density of 1 electron/m…

If R be the radius and σ the surface charge density of earth, then the total charge q on its surface is given by

Q = 4πR²σ

(i) The potential V at a point on earth's surface is same as if the entire charge q were concentrated as its centre. Thus

$V=\frac{1}{4 \pi \varepsilon_0} \frac{q}{R}=\frac{1}{4 \pi \varepsilon_0} \frac{4 \pi R^2 \sigma}{R}=\frac{R \sigma}{\varepsilon_0}$

Substituting the given value:

$V=\frac{6.4 \times 10^6 \times\left(-1.6 \times 10^{-19}\right)}{8.9 \times 10^{-12}}$

= − 0.115 N-m/C = −0.115 J/C = − 0.115Volt.