Answer
(a) 8 Sheets.
(b) $.045 m^2$
(c) See below.
Work Step by Step
(a) The capacitance between the two plates depends on the separated distance between the two plates and we could use equation 24.2 in the textbook to determine the distance $d$ in the form
\begin{equation}
d = \dfrac{K \epsilon_o A }{C} \tag{1}
\end{equation}
Let us substitute the values of $\epsilon_o, A, K$ and $C$ into equation (1) to get the value of $d$
\begin{align}
d = \dfrac{K\epsilon_o A V_{ab}}{Q} = \dfrac{3 (8.85 \times 10^{-12} \,\text{F/m}) (6.16 \times 10^{-2} \mathrm{~m^2})}{1.0 \times 10^{-12} \,\text{F}} = 1.64 \,\text{mm}
\end{align}
This is the total distance required to produce this capacitance. Hence, we divide this value of $d$ by the thickness of one sheet of aluminum to find the number of sheets $n$
$$n = \dfrac{d}{d_{sheet}} = \dfrac{1.64 \,\text{mm}}{0.20 \,\text{mm}} = \boxed{8 \,\text{sheets}}$$
(b) For the thickness $d$ = 12 mm the area will be given by solving equation (1) for $A$
\begin{align*}
A &= \dfrac{Cd }{K \epsilon_o }\\
&= \dfrac{(1.0 \times 10^{-12} \,\text{F})(12 \times 10^{-3} \,\text{m}) }{3 (8.85 \times 10^{-12} \,\text{F/m})}\\
&= \boxed{0.45 \mathrm{~m^2}}
\end{align*}
(c) The area is inversely proportional to the dielectric constant as shown by equation (1), therefore she needs a larger area because the dielectric constant of Teflon is smaller than the paper.
