Answer
(a) $W = 2.2 \times 10^{-6} \,\text{J}$
(b) $ V_{a} - V_{b} = 523 \,\text{V}$
(c) $E = 8.71 \times 10^{3} \,\text{V/m}$
Work Step by Step
(a) The work done is equal to the change in kinetic energies where $K_a = 0$
\begin{align*}
W &= \Delta K \\
& = K_b -K_a \\
&=K_b\\
&= \boxed{ 2.2 \times 10^{-6} \,\text{J}}
\end{align*}
(b) The potential of the starting point with respect to the end point is
$$V_a - V_b $$
And it is calcauted by
\begin{align*}
V_{a} - V_{b} &= \dfrac{W }{q} \\
&= \dfrac{2.20 \times 10^{-6} \,\text{J}}{4.20 \times 10^{-9} \,\text{C}}\\
&= \boxed{523 \,\text{V}}
\end{align*}
(c) The electric field is related to the difference in potential between the two points
\begin{gather*}
V_{a} - V_{b} = EL \\
E = \dfrac{ V_{a} - V_{b}}{L} \\
E = \dfrac{523 \,\text{V}}{0.06\,\text{m}}\\
\boxed{E = 8.71 \times 10^{3} \,\text{V/m}}
\end{gather*}