Seems like Saturn

Using the principle of superposition of electric fields we have: $\vec{E}_{P}=\vec{E}_{sphere}+\vec{E}_{ring}$ Taking as gaussian surface a sphere with radius $H$ for using Gauss's law : $\oint E_{sphere} ds = \frac{Q_{enclosed}}{\epsilon_{0}}$ $\Rightarrow E_{sphere} \cdot 4\pi H^2 = \frac{\rho \cdot \frac{4}{3} \pi r^3}{\epsilon_{0}}$ $\Rightarrow E_{sphere} = \frac{\rho \cdot r^3}{3H^2 \epsilon_{0}} \Rightarrow \vec{E}_{sphere}=\frac{\rho \cdot r^3}{3H^2 \epsilon_{0}}\hat{k}$ And by using the definition of electric field: $\vec{E}_{ring}=\frac{1}{4\pi \epsilon_{0}} \displaystyle \int_{0}^{2\pi} \frac{\lambda \cdot R d\theta}{(R^2+H^2)^{\frac{3}{2}}} (-R cos(\theta), -R sin(\theta), H)$ By the geometry of the ring, the electric field produced by it in $x$ and $y$ directions will be cancelled: $\vec{E}_{ring}=\frac{\lambda\cdot H}{4\pi \epsilon_{0}(R^2+H^2)^{\frac{3}{2}}}\displaystyle \int_{0}^{2\pi}d\theta \hat{k}$ $\vec{E}_{ring}=\frac{\lambda \cdot H}{2(R^2+H^2)^{\frac{3}{2}} \cdot \epsilon_{0}}$ So, as both of them has same direction, the magnitude will be: $E_{P}=\frac{\rho \cdot r^3}{3\cdot H^2 \cdot \epsilon_{0}}+\frac{\lambda \cdot R\cdot H}{2(R^2+H^2)^{\frac{3}{2}} \cdot \epsilon_{0}}$