Answer
(a) $E = 5.3 \times 10^{3} \mathrm{N} / \mathrm{C}$
(b) $V = 117 \mathrm{~V}$
(c) $E$ is the same while $V$ is doubled.
Work Step by Step
(a) The electric field between the two plates depends on the area of the plates and the charges on the plates in the form
\begin{gather}
E = \frac{Q}{A \epsilon_{o}}\\
E=\frac{\sigma}{ \epsilon_{o}}\\
E=\frac{47 \times 10^{-9}\mathrm{~C/m^2}}{8.85 \times 10^{-12}\,\text{F/m}}\\
\boxed{E =5.3 \times 10^{3} \mathrm{N} / \mathrm{C}}
\end{gather}
(b) The potential difference $V$ between the plates is related to the electric field by
\begin{equation}
V=Ed = (5.3 \times 10^{3} \mathrm{N} / \mathrm{C})(2.2 \times 10^{-2}\,\text{m})= \boxed{117 \mathrm{V}}
\end{equation}
(c) The electric field will be the $\textbf{same}$ as the surface charge density is the same. But the potential will be $\textbf{doubled}$ as it is directly proportional to the separated distance.
