Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the capacitor and the resistor are in parallel, they have the same voltage.
$$V_R=V_C$$
Hence,$I_R$ has the same direction of $\varepsilon_0$ while $I_C$ leads the $\varepsilon_0$ by an angle of $90^\circ$.
See the figure below.
The peak current in the resistor is given by
$$I_R=\dfrac{V_R}{R}=\dfrac{\varepsilon_0}{R}\tag 1$$
and the peak current in the capacitor is given by
$$I_C=\dfrac{V_C}{X_C}= \varepsilon_0\omega C\tag 2$$
$$\color{blue}{\bf [b]}$$
From the second figure below, we can see that
$$I=\sqrt{I_R^2+I_C^2}$$
Plug from (1) and (2),
$$I=\sqrt{\dfrac{\varepsilon_0^2}{R^2}+ \varepsilon_0^2\omega^2 C^2}$$
$$\boxed{I= \varepsilon_0 \sqrt{\dfrac{1}{R^2}+ \omega^2 C^2}}$$
