RL is just a bleed resistor to discharge C2. So, when someone disassembles the toaster, they don't get zapped.
To discharge to a safe level in about a minute: 27k.
Power = 160^2/27k = ~ 1W, this is probably too much power to waste. Maybe 100k?
Need to know the coil resistance to determine appropriate values for R1 and R2.
Edit 1: I wouldn't call J3 "AC in", to me that means line (mains). Andy is right, it must be a current sense. I will try again when I get a good schematic.
Edit 2: New guess as to what the overall circuit looks like.
The relay is probably either a 5V or 12V relay. If it is a 120VAC 1000W heater, then the heating element is about 14 ohms when hot. RL needs to be chosen to get the appropriate voltage to Q1-E. Somewhere around 0.4 to 1.0 ohm. Lots of assumptions made.
R1 needs to create enough base current to saturate the transistor, need to know the relay voltage and resistance. R2 = 10 * R1 is often a good choice.
Edit 3: Previous version can't be correct, too much current in the bridge rectifiers. Maybe the heater is split into 2 pieces. R11 produces just enough voltage to power the circuits.
To further analyze, we need to know the characteristics of Bridge_Out. Is it smooth and it is used to power the circuits? Then R6 (RL) is maybe 100 ohms. But, then Counter_Rst doesn't make sense.
The physical size of R6 (RL) might give some clues. If we knew the power, the we could estimate the resistance better.
If R11 opens, the circuit is toast LOL. Maybe this is what happened.
simulate this circuit – Schematic created using CircuitLab