1.2 mcd? Sheesh ... I knew there was something about blinkers I didn't like. You'll be doing good to see it unless you're standing in a dimly lit room.
OK, using the nominal values you listed, a forward drop of 2.5V across the LED and a 9V source (battery), you're looking at the resistor having to drop 6.5V (when the battery is fresh). Given a current drain of 55mA, ohm's law would give you a resistance of 118ohms ... 120 ohms is the closest standard value you'll find. Those crazy EEs ...
Seriously though, ohm's law (if you're interested) would work out like this:
R = E/I R = Resistance, E = Voltage, I = Current
But the voltage you use in the calculation isn't the voltage of the battery or what's dropped across the LED ... it's the voltage that's left over. What the resistor will have to drop, which is 9V (battery) - 2.5V (LED) = 6.5V (resistor). Take that voltage, divided by the current through the circuit (series circuit will have the same current flowing throughout), and there's the resiatance you need. As stated before, anything higher than this will ensure you're circuit is safe ... but will also give you a reduced output (dimmer).
Oh, and you'll be dropping about 0.35 watts across the resistor. So if you want a "bullet proof" circuit, you'll want a 1/2-watt resistor ... but since it's pulsed, you should be able to get away with an 1/4-watt.
Good luck,
ATM