= Solution
<Gauss law> states that for outward-oriented boundary $S=\partial V$,
$$
\boxed{\int_S\mathbf E\cdot\mathbf n\,dS=\frac1{\epsilon_0}\int_V\rho\,dV}.
$$
For a continuously differentiable field and <continuous> density, the <divergence theorem> turns the left side into $\int_V\nabla\cdot\mathbf E$. Thus the differential law implies the integral law. Conversely, if the integral law holds for every suitable volume, the integral of $\nabla\cdot\mathbf E-\rho/\epsilon_0$ vanishes on every small ball. A nonzero <continuous> value would keep one strict sign in some small ball, a contradiction. Hence $\nabla\cdot\mathbf E=\rho/\epsilon_0$. Weaker regularity gives the corresponding almost-everywhere or distributional formulation.
For the radial field, the flux on a radius-$r$ sphere is $4\pi r^2E(r)$, while its enclosed charge is $4\pi\int_0^r s^2\rho(s)\,ds$. Therefore
$$
\boxed{E(r)=\frac1{\epsilon_0r^2}\int_0^rs^2\rho(s)\,ds}.
$$
This uses the stated density as the full charge distribution, with no additional point charge at the origin. Since it vanishes beyond $a$, the total charge is
$$
\boxed{Q=4\pi\int_0^a s^2\rho(s)\,ds},
$$
and for $r>a$ the field is $\boxed{\mathbf E=Q\mathbf e_r/(4\pi\epsilon_0r^2)}$.
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