Courtesy of /u/Ame-No-Nobuko

Specific heat of water is 4.179 J/g˚C (http://www.iun.edu/~cpanhd/C101webnotes/matter-and-energy/specificheat.html)
Thus Gooperman! has a specific heat of 20895 J/g˚C
The formula that dictates the change in temp is ∆Q=mc∆T
where m is mass
Q is heat
c is specific heat
And T is temp in celcsisu
Gooperman melts at 1500˚K or 1226.85˚C
Assuming he starts at room temperature is at most 25˚C
So ∆T is 1201.85
Thus it can be modeled as ∆Q=2.563503075x10^7(m)
You can modify it to be based on time using 33.3 liters/ms
Or 49950 kg per second
The heat transfer would be Q=-kAdT/dx
Where Q is in joules, A is surface area of exposure, dT is change in temp and dx is basically thickness
As he is a cube his surface area would be a function of his volume.  So cubed root of (33000t), **where t is time is the length of one side, and surface area is cubed root of (33000t)**, squared times 6
Change in temp would basically just be the the temp in the interior of the sun, so 15000000˚C (https://www.space.com/17137-how-hot-is-the-sun.html)
Thickness would be cubed root of (33000t) divided by 2
Also K is the conducitivty which I'll assume is the same was water: 1.4 × 10-5
Using this model https://cdn.discordapp.com/attachments/378767485601841152/460161746511069194/Screen_Shot_2018-06-23_at_12.18.21_PM.png
Mind you this is a far more accurate model then what I calced before, but anything after 1.586×10^-11 seconds of full body contact (the heat would have to be around his entire body at 15 million ˚C) he'd die
For reference if the beam was only like 0.5 m^2 instead of what I calced it would be 3.4e-9 seconds to burn through
If you change the t on the right you can model it as two seperate times
Since the calc I just did assumed he started in the sun
https://cdn.discordapp.com/attachments/378767485601841152/460163598669578240/Screen_Shot_2018-06-23_at_12.25.43_PM.png
So for him to last 20 minutes in the sun it would take only 6.7e-7 seconds of growth

If you want a simpler, but less accurate model you can also just equate the change in water using ∆Q formula I listed earlier