Answer
(a) $V_B$ = 80 V
(b) $|E| = 100 \,\text{V/m}$ and its direction from $B$ to $A$
Work Step by Step
(a) For two instants at $A$ and $B$, we apply the conservation law
\begin{gather*}
K_{A} + U_{A} = K_{B} + U_{B} \\
0 + qV_{A} = K_{B} + qV_{B} \\
V_{A} = \dfrac{K_{B}}{q} + V_{B} \\
V_{B} = V_{A} - \dfrac{K_{B}}{q} \\
V_B = + 30.0 \,\text{V} - \dfrac{(3.00 \times 10^{-7} \,\text{J})}{(- 6.00 \times 10^{-9} \,\text{C})}\\
\boxed{V_B = + 80.0 \,\text{V}}
\end{gather*}
(b) The electric field is related to the difference in potential between the two points
\begin{gather*}
\Delta V = E L\\
V_{A} - V_{B} = EL
E = \dfrac{ V_{A} - V_{B}}{L} \\
E = \dfrac{30.0 \,\text{V} - 80.0 \,\text{V}}{0.500 \,\text{m}}\\
E = -100 \,\text{V/m}\\
\boxed{|E| = 100 \,\text{V/m}}
\end{gather*}
The direction of the electric field is from $B$ to $A$
