Motherboard RGB diode control and resistance?
The datasheet gives you the forward voltage of each LED at 20 mA and you have a 12 V supply. Your series resistor, therefore has to drop \$ 12 - V_f \$ volts at 20 mA. From Ohm's Law we can calculate the series resistance for each LED as

$$ R = \frac {V}{I} = \frac {12 - V_f}{0.02} $$

@Passerby is right that we should check the power dissipation in the resistors too. For example, your red LED calculation (in comments) gave a value of 500 Ω for a \$ V_f \$ of 2.0 V:

$$ P = \frac {V^2}{R} = \frac {(12-2)^2}{500} = \frac {10^2}{500} = \frac {100}{500} = 0.2~W $$

A 1/4 W resistor will suffice (although it will be hot to touch). Give it some breathing space.