I think I know how to cube it. Since the state space/possible values of bits can be represented as 2^(bits) what if for this problem you changed it to 2^(bits)^3 and then plug in 1536 to get 2^1536^3 = x. Since x had over 10^22 digits in it I am not going to post it all here but go see for for yourself by plugging the values into
http://www.std.com/~reinhold/BigNumCalc.html. If you take the (log2)x = 4608 and this represents the number of bits (probably wrong about this but don't know why).
Nitrodon I think I could do the same thing for the numbers you came up with as well. Convert 65536 to 16 bits and then take 24 bits x 16 bits = 384 bits^2. Now if you did the same thing as above you would get 2^384^2 = x (which is another really big number). Now take (log2)x = 768. Once again I probably messed up somewhere but I don't know where.