capacitive power supply sizing
I believe your \$C_2\$ is correct so I won't touch that one. 

Regarding \$C_1\$, we want it, on average, to push 220 mA through R1. I will do an approximation, assuming that the diodes are ideal. So you should be within 10% of the actual answer. 

The RMS value for a sine wave is \$\frac{A}{\sqrt2}\$ where \$A\$ is the amplitude of the sine wave. 

\$220\$ mA DC  \$\rightarrow 220\$ mA \$×\sqrt2≃311\$ mA AC

The maximum voltage across \$C_1\$, once we're in steady state mode will be \$120\$ V \$-2V_f-\frac{3.3}{2}\$ where \$V_f\$ is the forward voltage of the diodes. I'll assume 0.75 V. 

So we have an RMS current, a voltage and a frequency. 

We also know this: \$Q=I×S\$ and \$C=\frac{Q}{V}\$

Where \$Q\$ = charge, \$S\$ = time, \$I\$ = current, \$V\$ = voltage

In our case \$S = \frac{1}{120}\$ s, \$I = 311\$ mA, \$V = 120-2V_f-\frac{3.3}{2}=116.85\$ V

\$C=\frac{0.331×\frac{1}{120}}{116.85}=23.778\$ µF

*puts \$23.778\$ µF into the simulator*

Hmm, I've messed up somewhere, but I am on the right track at least. The current through \$C_1\$ is \$1×\sin(2\pi60t)\$ A (according to simulation). I'm no rocket scientist.. so let's just scale that 1 A down to 331 mA. 

\$C=\frac{0.331×\frac{1}{120}}{116.85}×0.331=7.37595\$ µF

*puts \$7.37595\$ µF into the simulator*

3.1 V across our 15 Ω load. Ehh, it was an approximation and an inverse rocket science. The error was \$\frac{3.3-3.1}{3.3}=6\%\$, less than 10% as I stated.

The reason for why it's not 100% correct is because there are dead time when the diodes are not active, and my approximation implied that there was no dead time. Which is why my approximation gave an answer that was less than 3.3 V.

I do not encourage you to mark this as the correct answer since this is just an approximation.  But hey, it beats 53 volts.