Answer
a) $5.14\;\rm mA$
b) $3.64\;\rm mA$
Work Step by Step
a) We know that the period of the rectified voltage is given by
$$T=\dfrac{1}{f}=\dfrac{1}{120}=\bf8.33\times10^{-3}\;\rm s$$
and the time constant of this circuit is given by
$$\tau=RC=(33\times 10^3)(28\times 10^{-6})=\bf0.924\;\rm s$$
Now it is obvious that the time circuit $\tau$ is much greater than the period time $\tau\gt \gt T$. This means that the voltage across the capacitor is constant during the cycle.
Thus, the average voltage is equal to the maximum voltage.
$$I_{avg}=\dfrac{V_{max}}{R} $$
we know that the maximum voltage is given by
$$V_{max}=\sqrt{2}v_{\rm rms} $$
So,
$$I_{avg}=\dfrac{\sqrt{2}v_{\rm rms}}{R} $$
Plugging the known;
$$I_{avg}=\dfrac{\sqrt{2}\times 120}{33\times 10^3} $$
$$I_{avg}= \color{red}{\bf 5.14\times10^{-3}}\;\rm A$$
________________________________
b)
Now we need to find the time constant of this circuit with the new capacitor.
$$\tau=RC=(33\times 10^3)(0.1\times 10^{-6})=\bf 3.3\times10^{-3}\;\rm s$$
In this case, the periodic time is greater than the time circuit $T\gt \tau$ by 2.5 times.
This means that the voltage goes up and down until it reaches zero voltage.
Therefore, the average voltage here is equal to the RMS voltage.
$$I_{avg}=\dfrac{V_{\rm rms}}{R} =\dfrac{120}{33\times10^3}$$
$$I_{avg}= \color{red}{\bf 3.64\times10^{-3}}\;\rm A$$
