Since we are getting picky here I need to point out the difference between power (measured in Watts) and energy (measured in Jules). 1W = 1 J/s. The power flux density (in W/m^2) is not the same as the energy absorbed (which will be in Jules).
The PFD at distance r is easy. It is simply the Sun’s total radiant flux (P=3.8E26 W) spread over an area which is the surface of a sphere of radius r so:
PFD = P / (4 pi r^2) [W/m^2]
If you put in 1 au (r = 1.5E11 m) you get a PFD of around 1.35 kW/m^2 which is our nice, familiar solar constant. If you put in the comet’s perihelion distance (8.55E8 m) you get the aforementioned 41 MW/m^2. That is indeed a lot!
As Paul points out the energy absorbed is much more complicated since it depends on the albedo. It also depends on how long you integrate over. Also, the comet’s nucleus will get very hot and so it will be radiating energy the other way. Depending on the thermal conductivity of the surface layer it might get close to thermal equilibrium with the photosphere. That is an exercise for the reader.
Whatever happens, it isn’t going to be nice. Add to that the tidal forces being that close to the Sun (they go as 1/r^3) and it is a big question whether it will survive. That will depend on how big and consolidated it is. We get a ringside seat since we’ll be able to see it as it passes through the field of view of space based coronagraphs.
What other Solar System objects do anything as exciting as this?