Since we know that V_{p(2) }= 20V_{rms }and that PIV stands for the peak inverse voltage, we have to find an answer that is in peak voltage. We also know that for a bridge rectifier (and for full wave rectifiers) that PIV = V_{p(2) }so all that we need to do is convert the 20V_{rms }to peak voltage:

V_{peak} = \(\sqrt(2)\)V_{rms -> }\(\sqrt(2)\)*20V_{rms} = 28.3V_{peak}