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As mentioned there is a formula to calculate capacitive reactance (capacitive impedance):

Xc= 1/ (2 π F C)

Thus the .68 uF value presents 4700 ohms impedance at the moment it conducts. It conducts at the 330V peaks. It lets through 70mA  (330/4700). As the circuit runs this is averaged to a level about right to light an LED (20 mA).

It's a basic  method which covers the main factors. You can do similar calculations to obtain an average of 100 mA which you desire in your project.