[MrFixr55]

My $0.02:
From the RCA Receiving Tube Manual, Cathode Current (Plate I + Screen I) for one tube is 37 mA or 0.037A. For 2 tubes, 0.074A. Grid Bias is -16.5V (Funny, these are similar specs to 2A5, 42, 6F6, etc.) So if P (Watts)=VI, then 16.5 (the voltage drop across the resistor) X 0.037A (the current for 2 tubes, and it may be less) = 1.23W. Total amplifier power is not the criteria, the voltage drop across the resistor X the current in the circuit determines the power dissipated across the resistor, and therefore the wattage required.

Current in a circuit is constant. The Amp likely delivers between 5 and 10 watts total. The total dissipation of the 2 tubes (250V X 0.074A) is 18W. Usable power is less than that.

In these calculations, I assumed both tubes are conducting. The current varies and should be 180 degrees out of phase. However, at these more severe conditions, the calculations should hold. Therefore a 10 Watt or even a 5 Watt resistor should handle it. I have a Magnavox amp that has 2 6V6s and plate voltage is about 300V. the common cathode resistor is a 2 watt carbon job.

Hope this helps.