No, this is the wrong calculation. The heat from the cylinder is much lower where it is surrounded by insulated foil that reflect back a good proportion of the heat. You might perhaps get a 20X amplification factor (taking extreme values) due to emissivity difference and insulation if that was very high. But that is still ball park only 5W.
Okay, so do the right calculation. Show us! You have 5 W left after the air removes most of the 50 W. Show us how that could all go out the hole, and fool two RTDs, a thermometer and two thermocouples.
Put a number on it. Do a quantitative analysis.
How the hell could it be 5 W reach a 5 cm hole 65 cm above the cylinder when 45 W are removed by the moving air? Does all of the heat bounce around, get reflected back, and then it all shoots right out that hole? Is the foil in the box in the shape of a parabola aimed at the hole? None of the heat radiates out from the wall? How would that work, anyway? Give us a model, instead of more hand waving, and baseless estimates of 5 W.