You have two 2.2k resistors, one in series with the LED, the other in series with the Triac gate supply. Which one do you mean? The forward current in the opto LED is probably about 10mA so you need R = (24V-1.8V-1V)/10mA giving a resistance in kOhms. 1.8V is the approximate visible LED forward drop and 1V is for the IR LED inside the opto. I don't have datasheets to hand for the parts you show but this topic covers questions about an almost identical requirement in great detail: [https://easyeda.com/forum/topic/Please-verify-my-AC-driver-circuit-using-OPTO-TRIAC-760863af4ae0452db1c4ae89d1f38a4c](https://easyeda.com/forum/topic/Please-verify-my-AC-driver-circuit-using-OPTO-TRIAC-760863af4ae0452db1c4ae89d1f38a4c) and should answer your question and may be some more you haven't thought of asking yet... :)

Sorry...  1.2k not 2.2k!

@andyfierman I meant the resistor R10, to trigger the gate of the triac. **Below I attached a picture of the original board I want to redesign.** 120V input on top left (we want to change it to 230V). Top middle is the connection to the lamps (Phase is switched, the lamps have a common neutral). Bottom connector is the 24V signals. The transformer and voltage regulator are used to make 5V, as the operating voltage for the triacs I think.  Triacs: 2N6071A Optocoupler:TIL113 Both are obsolete.