A solid insulating ball of radius 0.10 m holds 20 nC spread evenly through its volume…

Step 1. At 0.05 m, inside. $E = \dfrac{Qr}{4\pi\varepsilon_0 R^3} = \dfrac{20 \times 10^{-9}\ \text{C} \times 0.05\ \text{m}}{4\pi \times 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2) \times (0.10\ \text{m})^3} = 8.99 \times 10^3\ \text{N/C}$, that is $8.99$ kN/C.

Step 2. At the surface, 0.10 m. $E = \dfrac{Q}{4\pi\varepsilon_0 R^2} = \dfrac{20 \times 10^{-9}}{4\pi \times 8.85 \times 10^{-12} \times (0.10)^2} = 1.80 \times 10^4\ \text{N/C}$, that is $18.0$ kN/C.

Step 3. At 0.20 m, outside. The distance is twice as large, so the field is a quarter: $\dfrac{18.0}{4} = 4.50$ kN/C.

Step 4. The density. $V = \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi \times (0.10\ \text{m})^3 = 4.189 \times 10^{-3}\ \text{m}^3$. So $\rho = \dfrac{20 \times 10^{-9}\ \text{C}}{4.189 \times 10^{-3}\ \text{m}^3} = 4.77 \times 10^{-6}\ \text{C/m}^3$, that is $4.77\ \mu\text{C/m}^3$.