Answer
(a) $q = 2.78 \times 10^{-10} \mathrm{C} $
(b) The same charge $q = 2.78 \times 10^{-10} \mathrm{C} $
Work Step by Step
(a) From the force exerted we could get the charge $q$ by
$$F =qE$$
$$mg = q \frac{\sigma}{2 \epsilon_{c}}$$
Solving for $q$ we have
\begin{aligned}
q &=\frac{2\epsilon_{o} m g}{\sigma} \\
&=\frac{2\left(8.854 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}\right)\left(8 \times 10^{-6} \mathrm{kg}\right)\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)}{\left(5 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2}\right)} \\
&=2.78 \times 10^{-10} \mathrm{C}
\end{aligned}
(b) For the sheet, the electric field independent on the distance, so the charge is the same
$$q = 2.78 \times 10^{-10} \mathrm{C} $$
