You could use 6 LEDs, split into 2 groups of 3 LEDs in series with a resistor.
The max current is 30mA. Its a good idea to use less then the maximum value. It will make the LEDs last longer. When you get them (if you have a assortment of resitors around) you can play with them and see how bright they are at 30mA compared to 20mA or even less.
With the car off, the battery will be around 12v.
3 leds in series = Vf 3.5*3 = 10.5
12v - 10.5v = 1.5v
(Ohms law here)
20 mA = .020 A
R = V/I (V = IR rearranged)
R = 1.5/.020 = 75 ohm, should be a standard value
Wattage = V*I = .020 * 1.5 = .003 Watts - a 1/4 watt resistor will be fine
at 13.8v your current will rise to 44mA and the leds could burn out.
There are a couple ways around this.
Size the resistors for 13.8v
12v - 13.8v= 3.3v
20 mA = .020 A
R = V/I (V = IR rearranged)
R = 3.3/.020 = 165 ohm, they don't make a 165 ohm, but they make a 150 ohm
Replacing 165 with 150, If = 3.3/150 = .022 A (22 mA)
This will keep them safe with the engine running, but might make them dim with it off.
You could also use a voltage regulator set at 9v. This will keep the voltage stable running or not. You can redo the calculations using 2 leds in series and a 9v supply.
You could also use a resistor on each LED. This will make the swing in the amount of current less, but isn't as effecient.
With a 510 ohm resistor
13.8V = 20mA
12v = 16.7 mA
Sorry, i got a little long winded there. I am bored