Consider a charged sphere of radius $a$ and volume charge density $\rho$. Find the total work done to assemble this sphere, i.e total electric potential of this sphere. Use, $$U ={1\over 8\pi} \int_{\text{Entire field}} E^2 dv$$

I am using cgs unit instead of SI to get rid of pesky $k$ from my equations.

Consider a sphere of arbitrary volume $v$ less than the sphere in question.

Then I got $$v = \frac43\pi r^3 \iff r = \left({3v \over 4\pi}\right)^{1/3}$$

Since $\displaystyle E = {Q\over r^2} = {\rho v\over r^2} = {\rho v^{1/3}(4\pi)^{2/3}\over 3^{2/3}}$, I got $$E^2 = {\rho^2 v^{2/3}(4\pi)^{4/3}\over 3^{4/3}} $$

Plugging this in the given equation,

$$U = {1\over 8\pi}\int^{V_f}_0 {\rho^2 v^{2/3}(4\pi)^{4/3}\over 3^{4/3}} dv$$

Where $V_f$ is the volume of the sphere in the question.

Hence I got $$U = {1\over 10}\left({4\pi\over 3}\right)^{1/3}{\rho^2v^{5/3} }$$

Since $q^2 = (\rho v)^2$, where $q$ is the total charge of the sphere, $$U = {1\over 10} \left({4\pi\over 3}\right)^{1/3} {Q^2 \over v^{1/3}}$$

Substituting for $\displaystyle v = {4\over 3}\pi a^3$, I get,

$$U = {Q^2\over 10a}$$.

The given answer is $\displaystyle U = \frac35{Q^2 \over a}$.

I am missing a factor of $6$ in my answer. I think I messed up my limits on integral. Please help me.