Replacement for 18v 3w bulb

Yup - deep quotes do get mismatched. I did not write the section below.

To me it was clearly right. It assumes you know the resistance will change, but specifies the resistance used in the calculation at 48 ohms. But at least I finally understand your objection. It would have been better if he had said that the resistance would change, but that he was using the only known value we have.

As already stated, I took it differently - that the calculation was based on the only known resistance value. The math is correct using that value, and supports the correct conclusions that a 12 volt bulb will burn out if powered by 18 volts, and that a 24 volt bulb will be dim if run at 18 volts. The fact that the resistance changes is true - and not at all important in the answer to the OP, who doesn't know enough in any event to have that factor explained in the answer. Phil took a shortcut in the explanation, leaving out parts based on them not being needed, just as I did in stating that "a 12 volt bulb will burn out if powered by 18 volts". The statement does not specify an 18 volt source capable of sufficient current.

Right - Phil stated that a 24 ohm series resistor was needed.

Well, then we disagree, and you disagree with yourself. You stated that 24 ohms is correct. That figure was computed based on the effective resistance calculated at the rated voltage.

Great. If you expose the bulb to a voltage other than its rating, the resistance will change from the calculated value - that's a given. Build your voltage divider to keep the voltage that the bulb "sees" to its rating, based on the calculated effective resistance and let us know what you find.