The resistor in that case will be in series with the LED/LEDs so their voltage drop must be subtracted first.

i.e. if you have 2 red LEDs in series with your resistor, you must first subtract 2 x 1.7v (or there abouts) first.

Remainder is the voltage that will appear across the resistor.

If 12v supply, then 12 -(2 x 1.7v = 3.4v) = 8.6v.

You then need to calculate the resistor at 8.6v divided by whatever current through the LEDs (say 20mA)

So resistor is 8.6/0.20 = 430R.

As that is not a standard size, then go to 470R, you won't be able to see the difference.

Then the watage would be 8.6 (v) x 0.02 (A) = 0.172W .....so 1/4w resistor.