It looks the trigger current for class A GFCIs is 5mA.
So a resistor with a value of 120V/5mA = 24kΩ from one of the hot poles (~120V nom.) to ground should trigger the GFCI.
However 24 kΩ (5% or 10% tolerance) is not a common value, 22 kΩ and 27 kΩ are.
So a common 22 kΩ (5% or 10% tolerance) resistor should do it with a trigger current of 5.5 mA at 120V.
Power dissipation is V * I = 120V * 5.5mA = 0.66 Watt. Now since it's supposed to be on 'load' for only a fraction of a second, 1/4 watt should suffice. As long as the circuit works as it should and shuts off fast.
Keeping the button down if it doesn't work will fry that 1/4 watt resistor in short time. Use a 1/2 Watt or 1 Watt resistor if you plan to do this.
How GFCIs work.