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Looking at the wikipedia page on red dwarf stars[1], it appears a small star (M9V) might have 8% of the sun's radius and 0.015% of its luminosity. Thus, it would have about 156 times less area, and 2.3% of the output per area of the sun. So taking your figure as given, that means it would be less than 500W / cm^2.

[1]https://en.wikipedia.org/wiki/Red_dwarf

The temperature of the sun at its surface is ~5772 Kelvin. To get power per unit area, use Stefan-Boltzman: σ * T^4, σ * (5772 K)^4 ≈ 6294 W/cm^2. Dividing the sun's luminosity (power) by its surface area will also give a similar value.

A 815 mm^2, 250 W GPU will be 250 W / 8.15 cm^2 ≈ 31 W / cm^2.

Doesn't that only include radiated heat? If I touched it, it would be hotter.

It wouldn't be hotter (the temperature wouldn't change), but it would transfer more heat to you.

> 20kW of power per square centimeter

I'm getting 6300 W/cm^2 (see: https://news.ycombinator.com/item?id=17589371)