Answer :
To be able to solve this problem it is necessary to take into account the concepts of Electric Flow
on a closed surface.
By definition we know that the flow across a closed surface is
[tex]\Phi = \oint\vec{E}d\vec{A}[/tex]
And we know that the Coulumb's law says
[tex]\vec{E} = \frac{q}{4\pi\epsilon_0r^3}\vec{r}[/tex]
We can replace, so
[tex]\Phi = \oint \frac{q}{4\pi\epsilon_0r^3}\vec{r}d\vec{A}[/tex]
Solving the integral we reach the expression,
[tex]\Phi = \frac{q}{\epsilon_0}[/tex]
The charge of a proton is [tex]q=1.6*10^{-19}C[/tex], but we have 2 protons, then
[tex]q=(2)*1.6*10^{-19}C=3.2*10^{-19}C[/tex]
The constant of permittivity of free space [tex](\epsilon_0[/tex]) is [tex]8.85 x 10^{-12}[/tex]H/m
Replacing we have that the flux is
[tex]\Phi = \frac{q}{\epsilon_0}[/tex]
[tex]\Phi = \frac{3.2*10^{-19}}{8.85 x 10^{-12}}[/tex]
[tex]\Phi = 3.62*10^{-8}N m^2 C^{-1}[/tex]
We can also calculate the Volume of a Sphere, that is
[tex]V=\frac{4}{3} \pi r^3 =\frac{4}{3}\pi (1.9*10^{-15})^3[/tex]
[tex]V=2.8730*10^{-44}m^3[/tex]
Through this value is easy now calculate the dirvegence of the electric field, which is given by,
[tex]\xi = \frac{\Phi}{V}[/tex]
[tex]\xi = \frac{3.62*10^{-8}}{2.8730*10^{-44}}[/tex]
[tex]\xi = 1.2*10^36m^{-1} C^{-1}[/tex]