Answer
a) ${\bf 25}\;\rm mA$
b) ${\bf 6.63}\;\rm V$
Work Step by Step
$$\color{blue}{\bf [a]}$$
To find the peak current, we need to find the equivalent capacitance, as we did in the figure below.
$$I_{\rm peak}=\dfrac{\varepsilon_0}{X_C}=2\pi f C_{eq,2} \varepsilon_0$$
Plug the known (see the figure below);
$$I_{\rm peak}= 2\pi (200) (2\times 10^{-6}) (10)$$
$$I_{\rm peak}= \color{red}{\bf 25}\;\rm mA$$
$$\color{blue}{\bf [b]}$$
To find the peak voltage across the $3\;\mu$F capacitor, we need to use the second circuit below. It is obvious now that all the current passes through this capacitor.
So,
$$V_{1}=IX_{C1}=\dfrac{I}{2\pi f C_1}$$
Plug the known;
$$V_{1}= \dfrac{(25\times 10^{-3})}{2\pi (200)(3\times 10^{-6})}$$
$$V_{1}=\color{red}{\bf 6.63}\;\rm V$$
