Assuming that each GOW pulls 40mA (0.04A).

First you have to deduct the bulb voltage from the battery voltage. (6.4 volts minus 3.0 volts = 3.4 volts).

Now divide the remaining voltage by the current draw (3.4 volts divided by 0.04 amps = 85 ohms). The nearest common value would be 82 ohms (grey, red, black and gold bands).

One other thing to take into account, and that's the actual wattage rating of the resistor. The worse scenario being that the wattage value would be too low, would overheat, ending up open-circuit. So to calculate watts you simply multiply the volts by the amps. In this case, your 3-volt GOW multiplied by 0.04 amps = 0.12 watts (one-eighth watt). In reality you would need plenty of headroom, so best to multiply the wattage value by at least a factor of two (0.12 x 2 = 0.24 watts, or one-quarter watt). Two GOWs in parallel would draw 80mA, which in turns means the resistor would be best happy with a rating of at least 0.5 watts (one-half watt).

The reason you must be generous with the wattage rating is due to the heat factor (read wasted battery energy). For instance, if the resistor wattage value is underrated, it means the resistor itself will start to struggle, won't be able to cope and start to get hot. Which ultimately means a lot of the energy in your battery is being converted into wasted heat, rather than light.

You say that your floodlights are 6v bulbs. No problem. You no doubt have it figured by now that it's okay to stick the 6v lamps straight across the battery. (i.e: no dropper resistor needed). 2 x 3v bulbs in series is another way to do it (as you now know)

So the answer to your question is: Each series resistor would be 82-ohms, 1/4 watt.