Answer
a) $200 \;\rm N/C$
b) $101\;\rm N\cdot m^2/C$
c) $ 89\;\rm nC$
Work Step by Step
$$\color{blue}{\bf [a]}$$
Using the given formula
$$\vec E=5000r^2\;\hat r$$
Plugging the known,
$$\vec E=5000(0.2)^2\;\hat r$$
$$\vec E=\color{red}{\bf 200}\;\hat r\;\rm N/C$$
$$\color{blue}{\bf [b]}$$
We know that the electric flux for a closed surface is given by
$$\Phi=\oint \vec E\;d\vec A=EA_{\rm sphere}$$
where $A_{\rm sphere}=$
$$\Phi =4\pi r^2E$$
Plug the known;
$$\Phi =4\pi (0.2)^2(200)$$
$$\Phi =\color{red}{\bf 101}\;\rm N\cdot m^2/C$$
$$\color{blue}{\bf [c]}$$
Since the sphere is a closed surface, the electric flux through it is given by
$$\Phi=\dfrac{Q_{in}}{\epsilon_0}$$
Thus, the charge inside this sphere is given by
$$Q_{in}=\epsilon_0 \Phi$$
Plug the known;
$$Q_{in}=(8.85\times 10^{-12})(101)$$
$$Q_{in}=\color{red}{\bf 89.0}\;\rm nC$$
