OK, back of a fag packet calculation:

Energy in a cap is C(V*V)/2. Let's assume you could have an allowable drop from 3.3V to 2V for a full transmission, which lasts 10ms and takes a constant current of 100mA.

We'll give a very worst case energy used during transmission and say that's at constant 3.3V (voltage is dropping from 3.3V to 2V so this is worst case). Energy is power times time, so that's 3.3*100e-3*10e-3, i.e. 3.3mJ. That has to be equal to the loss of energy in the caps, so:

C*(3.3*3.3)/2 - C*(2*2)/2 = 3.3e-3.

C = (2*3.3e-3)/(3.3*3.3 - 2*2), which is roughtly 958uF. So you're looking at 1mF at least as a ball-park.

Mark.

Edit: Also remember that to charge this cap up you'd ideally want to to do it in an energy efficient manner. Just using a resistor will waste power in that resistor due to the I^2R losses. So some buck-boost regulator might be needed, but one that uses very low current (possibly even a switched capacitor regulator, although I have no real idea how efficient that might be). Unless of course you simply keep it charged up continually by the source and use a load switch to isolate it while transmitting..