Q: How to remove DC offset
When node B goes negative, the output of the op-amp will charge the capacitor via the diode. Feedback should keep the inverting input around 0 V. Not sure about stability of feedback loop. When node B goes positive, the output of the op-amp will be driven to ground but there is no feedback due to the diode. You could add a diode between the output of the op-amp and the inverting input for this half cycle but op-amp output will have to go slightly negative to maintain feedback. Perhaps you need a small negative supply or an offset input.