Answer
$W = 6.56 \times 10^{-21} \,\text{J}$
Work Step by Step
(a) The work done $W$ is the amount of energy applied on the charge for a distance $x$ where $x = 0.30 - 0.05 = 0.25 \,\text{m}$
$$W = Fx = qEx = qx \frac{\sigma}{2 \epsilon_{0}}$$
Substitute to get the work done
\begin{align}
W &= qx \frac{\sigma}{2 \epsilon_{0}}\\
&= \left(1.602 \times 10^{-19} \mathrm{C}\right) (0.25 \,\text{m}) \frac{2.9 \times 10^{-12} \mathrm{C} / \mathrm{m}^{2}}{2\left(8.854 \times 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)\right)}\\
&= \boxed{6.56 \times 10^{-21} \,\text{J}}
\end{align}
