Answer
a) $1.5\times 10^{-10}\;\rm F$
b) $12\;\rm kV$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the capacitance of a parallel plate capacitor is given by
$$C=\kappa C_0=\dfrac{\kappa \epsilon_0 A}{d} $$
where $\kappa$ is the dielectric constant.
Plug the known;
$$C= \dfrac{\kappa_{\rm Teflon} \epsilon_0 A}{d} =\dfrac{(2.1)(8.85\times 10^{-12})(0.04)^2}{(0.20\times 10^{-3})}$$
$$C=\color{red}{\bf 1.5\times 10^{-10}}\;\rm F$$
$$\color{blue}{\bf [b]}$$
The maximum potential difference between the plates is given by
$$E_{\rm max}=\dfrac{(\Delta V)_{\rm max}}{d}$$
where $E_{\rm max}$ is the maximum possible electric field in the capacitor before breakdown for the Teflon.
$$(\Delta V)_{\rm max}=E_{\rm max}d$$
Plug the known;
$$(\Delta V)_{\rm max}=(60\times 10^6)(0.2\times 10^{-3})$$
$$(\Delta V)_{\rm max}=\color{red}{\bf 12}\;\rm kV$$
