Answer
a) $200\;\rm V$
b) $0.63\;\rm nC$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that
$$ \Delta V_C=Ed$$
Plug the known;
$$ \Delta V_C = (1\times 10^5)(2\times 10^{-3})$$
$$\Delta V_C=\color{red}{\bf 200}\;\rm V$$
$$\color{blue}{\bf [b]}$$
The charge in one plate is equal in magnitude to the charge on the other plate. So we can find the charge in one plate to know how much charge is in each plate.
We know that the electric field inside the capacitor is given by
$$E=\dfrac{\eta}{\epsilon_0}$$
where $\eta=Q/A$
$$E=\dfrac{Q}{A\epsilon_0}$$
Solving for $Q$,
$$ Q = A\epsilon_0E $$
where $A$ here is $\pi r^2$
$$ Q = \pi \epsilon_0r^2 E $$
Plug the known;
$$ Q = \pi (8.85\times 10^{-12})(1.5\times 10^{-2})^2 (1\times 10^5) $$
$$Q=\color{red}{\bf 6.3\times 10^{-10}}\;\rm C$$
