Calculating capacitance required for a pulse duration of time T using a class AB amplifier
This answer is not based on seeing a circuit diagram. A circuit would help a lot.

If the load is to mid-rail of the power supply and you are driving a sinewave at 145.6 Vp-p, then the current calculation is this.


Calculate the RMS voltage - 145.6 Vp-p is 51.5 Vrms
Calculate the load power - \$\frac{V^2}{R}\$ = 3.96 watts
Estimate input power - input power is going to be no-worse than twice output power so say 8 watts max.
Estimate dc current drawn from each 120V - each half cycle occupies 50% of the time of the sine period therefore each power supply is supplying 4 watts or 33 mA.


Sanity check, 240V x 33mA = 8 watts. OK I think I got it right.

Your power supplies can supply 50mA and this might be OK without adding capacitance (there'll be some inside the power supplies anyway) and the peak current taken by the load is 

\$\sqrt2 \times 33mA\$ = 47mA.

I think you should be OK without adding capacitors and will run indefinitely as far as I can see.