Calculating required thermal resistance (rsa)
How is this not just a straight forward math problem?  Work it out:

You say the thermal resistance from junction to case is 4&deg;C/W, and that the junction is producing 10 W of heat.  It will therefore be 40&deg; hotter than the case.  The max allowed junction temperature is 150&deg;C, so the max allowed case temperature is 110&deg;C.

You say the worst case ambient temperature is 50&deg;C, so that means 10 W of heat flow from the case to ambient can't drop more than 60&deg;C.  Overall the case to ambient path must have (60&deg;C)/(10 W) = 6&deg;C/W thermal resistance or less.

Of course in practise you'd want to make sure the case to ambient resistance is some margin below the absolute maximum of 6&deg;C/W.