Answer
a) ${\bf8 0} \;\rm kHz$
b) ${\bf0}\;\rm V$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know, for a simple AC capacitor circuit, that
$$I_C=\dfrac{V_C}{X_C}$$
where $X_C=1/\omega C$, so
$$I_C= \omega C V_C $$
Recalling that $\omega=2\pi f$, so
$$I_C= 2\pi f C V_C $$
Solving for the frequency $f$,
$$f=\dfrac{I_C}{2\pi C V_C }$$
Plug the known;
$$f=\dfrac{(50\times 10^{-3})}{2\pi (20\times 10^{-9})(5) }$$
$$f\approx \color{red}{\bf8 0} \;\rm kHz$$
$$\color{blue}{\bf [b]}$$
We know, for a simple AC capacitor circuit, that
$$i_C=I_C\cos(\omega t+\frac{\pi}{2})$$
the $\pi/2$ is because the current leads the voltage through a capacitor by $\pi/2$
To have $i_C=I_C$, we must have $\cos(\omega t+\frac{\pi}{2})=1$
This occurs when $(\omega t+\frac{\pi}{2})=2\pi$ and for general cases, $2\pi n$ where $n=1,2,3,...$.
Thus,
$$\omega t=2\pi n-\dfrac{\pi}{2}=\dfrac{4\pi n-\pi}{2}=\dfrac{4n-1}{2}\pi$$
Hence, for $n=1,2,3,...$,
$$\omega t= \frac{3}{2}\pi, \frac{7}{2}\pi, \frac{11}{2}\pi,...$$
Plug these values into
$$v_C=V_C\cos(\omega t)=0\;\rm V$$
This is zero for all $\omega t$ values.
So that the instantaneous value of the emf at the instant when $i_C=I_C$ is zero volt.
$$v_C=\color{red}{\bf0}\;\rm V$$
