= Solution
In SI units, <Gauss's law> for a closed surface with outward unit <normal vector> is
$$
\oint_S\mathbf E\cdot\mathbf n\,dS=\frac1{\varepsilon_0}\int_{\operatorname{inside}(S)}\rho\,dV=Q_{\rm enclosed}/\varepsilon_0.
$$
Spherical symmetry and the absence of an imposed external field make the <electric field> radial, $\mathbf E=E_r(r)\mathbf e_r$. For $0\leq r\leq a$ (with the physical radius $a>0$),
$$
Q(r)=4\pi K\int_0^r(a^2s^2-s^4)\,ds=4\pi K\left(\frac{a^2r^3}{3}-\frac{r^5}{5}\right).
$$
Outside the charge distribution this becomes $Q(a)=8\pi K a^5/15$. Applying <Gauss's law> on spheres gives
$$
\boxed{\mathbf E(\mathbf x)=\begin{cases}\displaystyle\frac K{\varepsilon_0}\left(\frac{a^2r}{3}-\frac{r^3}{5}\right)\mathbf e_r,&0<r\leq a,\\[3pt]\displaystyle\frac{2Ka^5}{15\varepsilon_0 r^2}\mathbf e_r,&r\geq a,\\0,&r=0.\end{cases}}
$$
Both formulas agree at $r=a$, and the first tends to zero at the origin. Thus \b[the total charge is $8\pi K a^5/15$]. Negative $K$ reverses the field direction. If the radius is zero the distribution vanishes; a negative value of $a$ gives no nonnegative-radius support under the printed piecewise inequalities and likewise gives the zero distribution.
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