Answer
(a) $q = -3.21 \times 10^{5} \mathrm{C}$
(b)$E = 250 \mathrm{N} / \mathrm{C} $
(c) $\sigma = -2.21 \times 10^{-9} \mathrm{C} / \mathrm{m}^{2} $
Work Step by Step
(a) The charge from the electric flux is calculated by
\begin{aligned}
q &= \Phi_{E}\epsilon_{o} \\
&=\left(-3.63 \times 10^{16} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\right)\left(8.854 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}\right) \\
&=\boxed{-3.21 \times 10^{5} \mathrm{C}}
\end{aligned}
(b) The electric field is found by knowing the area
\begin{aligned}
E &=\frac{\left|\Phi_{E}\right|}{4 \pi R^{2}} \\
&=\frac{\left(- 3.63 \times 10^{16} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\right)}{4 \pi\left(3.40 \times 10^{6} \mathrm{m}\right)^{2}} \\ &=250 \mathrm{N} / \mathrm{C}
\end{aligned}
(c) The charge density is the charge per unit area
\begin{aligned}
\sigma &=\frac{q}{4 \pi R^{2}} \\
&=\frac{-3.21 \times 10^{5} \mathrm{C}}{4 \pi\left(3.40 \times 10^{6} \mathrm{m}\right)^{2}} \\
&=-2.21 \times 10^{-9} \mathrm{C} / \mathrm{m}^{2}
\end{aligned}
