You could put 2 or 3 in series. 2 would need about 6.6 V, and 3 would need about 9.9 V.
You should allow for 14 V with the engine running, so the voltage remaining is either 7.4 V or 4.1 V so the resistor you should use is 24 Ω for 2 LEDs or 14 Ω for 3 LEDs. The power rating is 2.5 W or more for the 2 LEDs and 1.5 W or more for 3 LEDs.
Now when the voltage falls to 12 V when the engine stops, the current will fall more if there are 3 LEDs in series. For 2 LEDs, the current will be (12 - 6.6)/24 = 225 mA. For 3 LEDs, the current will be (12 - 9.9)/14 = 150 mA
So if you have 2 LEDs in series, you need 25 sets of 2 in series so you need 25 resistors of 24 Ω, and the total current will be 7.5 amps. The current will drop to 3/4 when the engine stops.
If you have 3 LEDs in series, you will have 16 sets of 3 in series, and one set of 2 in series.
You will need 16 resistors of 14 Ω and one of 24 Ω and the total current will be 5.1 amps. The current will drop to 1/2 when the engine stops.
The alternative is a dedicated LED driver IC, which needs an inductor. Something like this
https://www.diodes.com/datasheets/AP8801.pdf would let you run 3 LEDs in series with no dimming at 12 V.
There are also boost converters that let you put more LEDs in series.