Answer
(a) $r = 0.30\,\text{m} $
(b) $q = 16.6 \times 10^{-11}\,\text{C}$
(c) Away.
Work Step by Step
(a) The electric field $E$ is related to the potential difference and the distance $r$ by
\begin{gather*}
E = \dfrac{V}{r}\\
r = \dfrac{V}{E}\\
r = \dfrac{4.98 \,\text{V}}{16.2 \,\text{V/m}} \\
\boxed{r = 0.30 \,\text{m} }
\end{gather*}
(b) The potential difference is related to the charge $q$ by
\begin{gather*}
V = \dfrac{1}{4 \pi \epsilon_{\circ}} \dfrac{q}{r} \\
q = 4 \pi \epsilon_{\circ} rV \\
q = 4 \pi \epsilon_{\circ} rV \\
q = \dfrac{1}{9.0 \times 10^{9} \mathrm{N\cdot m^{2}/C^{2}}} (0.30 \,\text{m})(4.98 \,\text{V})\\
\boxed{q = 16.6 \times 10^{-11}\,\text{C}}
\end{gather*}
(c) The electric field is $\textbf{away}$ from the charge as it is positive.
