Independent Electric Field

Infinitesimal charge:

$dQ = \rho \, dV = \frac{\alpha}{r} \, 4 \pi \, r^2 \, dr = 4 \pi \, \alpha \, r \, dr$

Integrating gives the total charge out to radius $r$ :

$Q(r) = 2 \pi \, \alpha \, r^2$

Since the charge distribution has radial symmetry, we can use the Gauss Law:

$E(r) = \frac{Q(r)}{4 \pi \, r^2 \, \epsilon_0} = \frac{2 \pi \, r^2 \, \alpha}{4 \pi \, r^2 \, \epsilon_0} = \frac{1}{2} \frac{\alpha}{\epsilon_0}$