Answer
a) ${\bf50}\;\rm Hz$
b) ${\bf 4.77}\;\rm \mu F$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the frequency is given by
$$f=\dfrac{1}{T}$$
where, from the given graphs, the period is 0.02 s.
$$f=\dfrac{1}{0.02}=\color{red}{\bf50}\;\rm Hz$$
$$\color{blue}{\bf [b]}$$
Recalling that the voltage across the capacitor is given by
$$V_C=I_CX_C$$
where $X_C=1/\omega C$, so
$$V_C=\dfrac{I_C }{\omega C}$$
Hence the capacitance is given by
$$C=\dfrac{I_C }{\omega V_C }=\dfrac{I_C }{2\pi f V_C }$$
We can see, from the given graph, $I_C=15$ mA, and $V_C=10$ V.
Plug the known;
$$C =\dfrac{(15\times 10^{-3})}{2\pi (50)(10) }$$
$$C =\color{red}{\bf 4.77}\;\rm \mu F$$
