Assuming for a moment that a simple solution is allowed, and you don't need to worry about up, down, etc, you've got yourself a box .2 meters to a side, or .24 sq.m. in area form .2^2x6=.24 ... this is at equilibrium with a presumed infinite heat reservoir of 60C outside, meaning you are losing 200 W per .24 m or 833.33 W/m^2 and your insulator is .01 m thick.

Now assuming my brain hasn't fried from not doing this for years, the formula is kAT/x =Q/t where Q/t should be 833.33 W (for a square meter) and kAT/x where k is what we want an A is area and T is 60C-20C and x is 1 cm or .01m should be k(1)(40)/ (.01) so that 833=k40./.01 or k=833/4000=.2083 Watts per meter*kelvin or .2 W/mK which is fairly sane value so that's probably right, but I'd let one of the other physicists on the site confirm that since we have quite a few and as mentioned I haven't done it in a while

and redoing that where we just go with Q/t=200W=kAT/x would be =k(.24m2)(60-20C)/ (.01m)= 200W*.01m/ [(60-20C)*.24m2]=k=.2W/m / 40C*.24 = 2W/m / 9.6 Km2 = .2083 W/Km if that's a bit more intuitive and step by step, of course it may still be wrong, check with someone else.