Answer
(a) $E = \dfrac{Q}{\epsilon_o K A} $
(b) $V = \dfrac{Q}{\epsilon_o K A} d$
(c) $C = \dfrac{\epsilon_o K A}{d}$
Work Step by Step
(a) We integrate equation 24.23 for $A$ to determine the electric field by
\begin{gather*}
\oint K \vec{E} \cdot d\vec{A} = \dfrac{Q}{\epsilon_o} \\
K E A = \dfrac{Q}{\epsilon_o} \\
\boxed{E = \dfrac{Q}{\epsilon_o K A} }
\end{gather*}
(b) We use the expression of the electric field to determine the potential $V$ by
$$V = E d = \boxed{\dfrac{Q}{\epsilon_o K A} d}$$
(c) The capacitance depends on the charge and the potential and it is given by
\begin{gather*}
C = \dfrac{Q}{V}\\
C = \dfrac{Q}{ \dfrac{Q}{\epsilon_o K A} d }\\
\boxed{C = \dfrac{\epsilon_o K A}{d}}
\end{gather*}
This is the same for equation 24.2 but after inserting a dielectric material
