(A) Let the Universe have a radius $R$. Assume that the hydrogen atoms are uniformly distributed. The expansion of the universe will start if the Coulomb repulsion on a hydrogen atom at $R$ is larger than the gravitational attraction.
The hydrogen atom contains one proton and one electron. The net charge on each hydrogen atom is:
$q = e_p + e = -(1 + y)e + e = -e - ye + e = -ye$.
Let $E$ be the electric field intensity at distance $R$ on the surface of the sphere. According to Gauss's theorem:
$\oint \vec{E} \cdot d\vec{S} = \frac{q_{enclosed}}{\epsilon_0}$
$E(4\pi R^2) = \frac{1}{\epsilon_0} \left( \frac{4}{3} \pi R^3 N |ye| \right)$
$E = \frac{N|ye|R}{3\epsilon_0}$.
The Coulomb force on a hydrogen atom at $R$ is:
$F_C = qE = (ye) \left( \frac{NyeR}{3\epsilon_0} \right) = \frac{y^2 e^2 N R}{3\epsilon_0}$.
The gravitational field $G_R$ at distance $R$ is given by:
$G_R = \frac{4}{3} \pi G m_p N R$.
The gravitational force on the atom is:
$F_G = m_p G_R = \frac{4}{3} \pi G m_p^2 N R$.
Expansion starts when $F_C > F_G$:
$\frac{y^2 e^2 N R}{3\epsilon_0} > \frac{4}{3} \pi G m_p^2 N R$
$y^2 > \frac{4 \pi G m_p^2 \epsilon_0}{e^2}$.
Thus,the critical value is $y = \sqrt{\frac{4 \pi G m_p^2 \epsilon_0}{e^2}}$.
$(b)$ The net force on the atom is $F_{net} = F_C - F_G = kR$,where $k = \frac{y^2 e^2 N}{3\epsilon_0} - \frac{4}{3} \pi G m_p^2 N$. Since $F = ma$,the acceleration $a = \frac{k}{m_p} R$. Since $a = \frac{dv}{dt} = v \frac{dv}{dR}$,integrating $v dv = \frac{k}{m_p} R dR$ gives $v^2 \propto R^2$,so $v \propto R$.