Answer
a) $1.77\times10^{-2}\;\rm A$
b) $-12.6^\circ$
c) $117\;\rm V, \;26.1\;V$
Work Step by Step
a) We know that
$$I_{\rm rms}=\dfrac{V_{\rm rms}}{Z}\tag 1$$
we are given the arms voltage $V_{\rm rms}$, so we need to find the impedance $Z$.
We know that the impedance is given by
$$Z=\sqrt{R^2+(X_L-X_C)^2}$$
And in this case, we have $RC$-circuit without an inductive reactance$X_L=0$.
Thus,
$$Z=\sqrt{R^2+ X_C^2}=\sqrt{R^2+ \left(\dfrac{1}{2\pi f C}\right)^2}$$
Plugging into (1);
$$I_{\rm rms}=\dfrac{V_{\rm rms}}{\sqrt{R^2+ \left(\dfrac{1}{2\pi f C}\right)^2}} $$
Plugging the known;
$$I_{\rm rms}=\dfrac{ 120}{\sqrt{(6.6\times10^3)^2+ \left(\dfrac{1}{2\pi\cdot 60\cdot (1.8\times10^{-6})}\right)^2}} $$
$$I_{\rm rms}=\color{red}{\bf1.77\times10^{-2}}\;\rm A$$
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b) We know that the phase angle is given by
$$\phi=\tan^{-1}\left(\dfrac{\overbrace{X_L}^{0}-X_C}{R}\right)=\tan^{-1}\left(\dfrac{ -X_C}{R}\right)$$
$$\phi =\tan^{-1}\left(\dfrac{ -\dfrac{1}{2\pi f C}}{R}\right)=\tan^{-1}\left(\dfrac{ -1}{2\pi f CR}\right)$$
Plugging the known;
$$\phi= \tan^{-1}\left(\dfrac{ -1}{2\pi \cdot 60\cdot 1.8\times10^{-6}\cdot 6.6\times10^3 }\right)$$
$$\phi=\color{red}{\bf -12.6^\circ}$$
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c)
The voltage reading across the resistance is given by
$$V_R=I_{\rm rms}R=1.77\times10^{-2}\cdot 6.6\times10^3$$
$$V_R=\color{red}{\bf117}\;\rm V$$
The voltage reading across the capacitor is given by
$$V_C=I_{\rm rms}X_C=I_{\rm rms}\;\dfrac{1}{2\pi f C}$$
$$V_C= \dfrac{I_{\rm rms}}{2\pi f C}=\dfrac{1.77\times10^{-2}}{2\pi \cdot 60\cdot 1.8\times10^{-6}}$$
$$V_C=\color{red}{\bf 26.1}\;\rm V$$
