Answer
(a) $C = 5.3 \times 10^{-13} \,\text{F}$
(b) $\Delta d = 0.23 \,\text{mm} $
Work Step by Step
(a) The capacitance depends on the area $A$ of the plate and the distance between the two plates and it is given by
$$C = \dfrac{\epsilon_o A}{d}$$
Substitute by the values of $\epsilon_o, A$ and $d$ to get the capacitance $c$
\begin{align*}
C &= \dfrac{\epsilon_o A}{d} = \dfrac{(8.85 \times 10^{-12} \,\text{F/m}) (42\times 10^{-6}\mathrm{~m^2})}{0.7 \times 10^{-3}\,\text{m}} = \boxed{5.3 \times 10^{-13} \,\text{F}}
\end{align*}
(b) The area is constant, therefore, the change in the distance $d$ is related to the change in capacitance by
$$\Delta d +d = \dfrac{\epsilon_o A}{\Delta C + C} = \dfrac{(8.85 \times 10^{-12} \,\text{F/m} ) (42\times 10^{-6}\mathrm{~m^2})}{0.250 \times 10^{-12} \,\text{F}+5.3 \times 10^{-13} \,\text{F}} = 0.47 \,\text{mm} $$
The change in distance is
$$\Delta d =0.47 \,\text{mm} - d = 0.47 \,\text{mm} - 0.7 \,\text{mm} = -0.23 \,\text{mm} $$
The negative sign here means the decreasing of the separated distance where
$$\Delta d = 0.23 \,\text{mm} $$
