I understand that the bigger resistor is the lower current will but I miss some explanation about how to choose the resistor's value.
You have to make sure you only have Vf available to the LED, and need to lose any other voltage from the supply, across the resistor.
You know a few things: the voltage supplied Vs, the voltage the LED can take Vf, and the current I.
You need to lose the difference in the voltages across the resistor, call that Vr: so Vr = Vs - Vf
Apply Ohm's Law to the resistor, where it's R we need to find: R = Vr / I = (Vs - Vf) / I.
I don't know Vs in your example, let's say it's 9v for example... then we need to lose 9 - 2.6 = 6.4v across the resistor so R = 6.4 / .02 = 320.
So to summarise, you need to apply Ohm's Law to the resistor, not the LED, and the resistor needs to drop the surplus voltage.