To calculate the RMS voltage after rectification and smoothing, you need to consider the peak voltage of the AC input and the characteristics of the rectifier and smoothing capacitor. Given that you have a 19 volts RMS AC input, after full-wave rectification, the peak voltage will be √2 times the RMS voltage, which is approximately 1.414 times the RMS voltage. So, for a 19 volts RMS input, the peak voltage would be approximately 19 * 1.414 = 26.866 volts.
After rectification, you would have pulsating DC with peaks at this voltage. Then, after smoothing using a reservoir capacitor, the voltage will increase slightly above the peak voltage, depending on the capacitor's value and the load. But, generally, for no load conditions, it's safe to approximate that the output voltage will be close to the peak voltage. So, in your case, with a 19 volts RMS AC input, after rectification and smoothing at no load, you can expect an output voltage close to 26.866 volts DC.