Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The ball’s volume charge density is given by
$$\rho_e=\dfrac{Q}{V}=\dfrac{Q}{\frac{4}{3}\pi R^3}$$
Plug the known;
$$\rho_e =\dfrac{80\times 10^{-9}}{\frac{4}{3}\pi (0.2)^3}$$
$$\rho_e =\color{red}{\bf 2.39\times 10^{-6}}\;\rm C/m^3$$
$$\color{blue}{\bf [b]}$$
To find the charge enclosed at $r=5$ cm,
$$\rho_e=\dfrac{Q_1}{V_1}=\dfrac{Q}{V}$$
Hence,
$$\boxed{Q_1=\rho_e V=\frac{4}{3}\pi r_1^3 \rho_e}$$
Plug the known;
$$Q_1 =\frac{4}{3}\pi (0.05)^3 (2.39\times 10^{-6})$$
$$Q_1=\color{red}{\bf 1.25}\;\rm nC$$
when $r=10$ cm, using the boxed formula above,
$$Q_2 =\frac{4}{3}\pi r_2^3 \rho_e$$
Plug the known;
$$Q_2 =\frac{4}{3}\pi (0.1)^3 (2.39\times 10^{-6})$$
$$Q_2=\color{red}{\bf 10}\;\rm nC$$
when $r=20$ cm, using the boxed formula above,
$$Q_3 =\frac{4}{3}\pi r_3^3 \rho_e$$
Plug the known;
$$Q_3 =\frac{4}{3}\pi (0.2)^3 (2.39\times 10^{-6})$$
$$Q_3=\color{red}{\bf 80}\;\rm nC$$
$$\color{blue}{\bf [c]}$$
We know, from Gauss’s law, that
$$\Phi=\oint \vec Ed\vec A=\dfrac{Q_{in}}{\epsilon_0}$$
Thus,
$$EA=\dfrac{Q_{in}}{\epsilon_0}$$
$$E=\dfrac{Q_{in}}{\epsilon_0A}$$
where $A=4\pi r^2$ which is an area of a Gaussian sphere,
$$E=\dfrac{Q_{in}}{4\pi \epsilon_0 r^2 }$$
For $r_1=5$ cm,
$$E_1=\dfrac{Q_{1,in}}{4\pi \epsilon_0 r_1^2 }$$
Plug the known,
$$E_1=\dfrac{1.25\times 10^{-9}}{4\pi (8.85\times 10^{-12})(0.05)^2 }$$
$$E_1=\color{magenta}{\bf 4496} \;\rm N/C$$
For $r_2=10$ cm,
$$E_2=\dfrac{Q_{2,in}}{4\pi \epsilon_0 r_2^2 }$$
Plug the known,
$$E_2=\dfrac{10\times 10^{-9}}{4\pi (8.85\times 10^{-12})(0.1)^2 }$$
$$E_2=\color{magenta}{\bf 8992} \;\rm N/C$$
For $r_2=20$ cm,
$$E_3=\dfrac{Q_{3,in}}{4\pi \epsilon_0 r_3^2 }$$
Plug the known,
$$E_3=\dfrac{80\times 10^{-9}}{4\pi (8.85\times 10^{-12})(0.2)^2 }$$
$$E_3=\color{magenta}{\bf 17984} \;\rm N/C$$
