The question is contradictory : It says that the charge density is uniform (the same at every point in the hemisphere), but also that it depends on azimuthal co-ordinate $\phi$, with $\rho=3\phi$. The latter is a very unusual variation, because it has a discontinuity at the meridian $\phi=0, 2\pi$. Perhaps a polar variation is intended, with $\rho=3\theta$, which has no discontinuity.
(a) Uniform Volume Charge Density $\rho=\text{const}$
Divide the hemisphere into rings (hoops) of thickness $dr$ and width $rd\theta$ located at constant values of $r, \theta$ parallel to the xy plane, with the centre as origin. The radius of the ring is $y=r\sin\theta$, its volume is $dV=2\pi y r d\theta dr$ and its charge is $dq=\rho dV$.
The z-component of the electric field due to this ring at the origin is $dE_z=-kdq \cos\theta/r^2$. The total electric field due to all rings in the hemisphere is $$E_z=-\int dE_z=-k\pi \rho \int_0^a dr \int_0^{\pi/2} \sin2\theta d\theta =-\frac{\rho a}{4\epsilon_0} $$
The total charge on the hemisphere is $Q=\frac23 \pi \rho a^3$ so the electric field can be re-written as $$E_z=-\frac{3Q}{8\pi \epsilon_0 a^2}$$
(b) Polar Charge Density $\rho=3\theta$
Again we can use rings because they have a constant value of $\theta$.
$$E_z=-\int 3\theta dV=-k\pi \int_0^a dr \int_0^{\pi/2} 3\theta \sin2\theta d\theta =-\frac{a}{4\epsilon_0} \frac34 [\sin2\theta-2\theta\cos2\theta]_0^{\pi/2} =-\frac{3\pi a}{16\epsilon_0}$$
The total charge on the hemisphere is $$Q=\int dq=\pi\int_0^a 3r^2 dr \int_0^{\pi/2} \theta\sin2\theta.d\theta=\frac14 \pi^2 a^3$$ so the electric field can be re-written as $$E_z=-\frac{3Q}{4\pi \epsilon_0 a^2}$$
(c) Azimuthal Charge Density $\rho=3\phi$
We can no longer use rings because $\phi$ is not constant for a ring. The volume element is $dV=r^2\sin\theta dr d\theta d\phi$ and the charge on each element is $dq=3\phi dV$.
The electric field at the centre due to an element of charge at ($r,\theta,\phi$) has components $$dE_z=-k\frac{dq}{r^2}\cos\theta, dE_y=-k\frac{dq}{r^2}\sin\theta\cos\phi, E_x=-k\frac{dq}{r^2}\sin\theta\sin\phi$$ so we get $$E_z=-3k\int_0^a dr \int_0^{\pi/2} \frac12 \sin2\theta d\theta \int_0^{2\pi} \phi d\phi=-\frac{3\pi a}{2\epsilon_0}$$ $$E_y=-3k\int_0^a dr \int_0^{\pi/2} \frac12 (1-\cos2\theta) d\theta \int_0^{2\pi} \phi \cos\phi d\phi=-\frac{3 a}{2\epsilon_0}(\frac{\pi}{4}+1)$$ $$E_x=-3k\int_0^a dr \int_0^{\pi/2} \frac12 (1-\cos2\theta) d\theta \int_0^{2\pi} \phi \sin\phi d\phi=-\frac{3a}{2\epsilon_0}(\frac{\pi}{4}+1)$$
The total charge on the hemisphere is $$Q=\int dq=\int_0^a 3r^2 dr \int_0^{\pi/2} \sin\theta d\theta \int_0^{2\pi} \phi d\phi=2\pi^2 a^3$$ so we can express the components of the field as $$E_z=-\frac{3Q}{4\pi \epsilon_0 a^2}, E_y=E_x=-\frac{3Q}{4\pi \epsilon_0 a^2}(\frac14+\frac{1}{\pi})$$
