So the question is really to
count the maximal number of cubelets in transparent mode.
If there are $n$ of them, then the power needed is $4\times1000+n\times100+(121-n)\times10=5210+90\times n$ Watts.
A lower and upper bound for
this maximal number is 18 and 66 respectively.
The lower bound comes from an actual setup where the interferometer cubes are in 4 corners of the large cub, so they are pairwise in opposite corners of a side of the cube. Any two of these have 3 cubelets between them, different pair of interferometer cubes on different sides of the cube, so those are distinct cubelets, that's $6\times3=18$ in total.