Answer
(a) $E = 2.82 \times 10^{5} \mathrm{N} / \mathrm{C} $
(b) $E = 3.95 \times 10^{5} \mathrm{N} / \mathrm{C}$
(c) $E = 1.69 \times 10^{5} \mathrm{N} / \mathrm{C}$
Work Step by Step
(a) The electric field at point $A$ is the summation of the electric fluxes
\begin{align*}
E &=\frac{1}{2 \epsilon_{0}}\left(5 \mathrm{C} / \mathrm{m}^{2}+2 \mathrm{C} / \mathrm{m}^{2}+4 \mathrm{C} / \mathrm{m}^{2}-6 \mathrm{C} / \mathrm{m}^{2}\right)\times 10^{-6}\\
& =\boxed{2.82 \times 10^{5} \mathrm{N} / \mathrm{C} }
\end{align*}
(b) The electric field at point $B$ is the summation of the electric flux but each surface charge will be in the opposite sign
\begin{align}
E&=\frac{1}{2 \epsilon_{0}}\left(6 \mathrm{C} / \mathrm{m}^{2}+2 \mathrm{C} / \mathrm{m}^{2}+4 \mathrm{C} / \mathrm{m}^{2}-5 \mathrm{C} / \mathrm{m}^{2}\right)\times 10^{-6}\\
& =\boxed{3.95 \times 10^{5} \mathrm{N} / \mathrm{C}}
\end{align}
(c) At point $C$, the electric field is given by
\begin{align}
E &=\frac{1}{2 \epsilon_{0}}\left(4 \mathrm{C} / \mathrm{m}^{2}+6 \mathrm{C} / \mathrm{m}^{2}-5 \mathrm{C} / \mathrm{m}^{2}-2 \mathrm{C} / \mathrm{m}^{2}\right)\times 10^{-6}\\
& =\boxed{1.69 \times 10^{5} \mathrm{N} / \mathrm{C}}
\end{align}
