Thats the genius of the written language, you can look it up! Not a simple technique to get bond lengths, you can get very rough estimates from techniques like IR, but that requires a calibration curve formed from similar species.
The most accurate techniques are Neutron and Electron diffraction techniques. As far as I know the only place in the UK where this is possible is at the Diamond light source in Oxfordshire. (So a very, well not necessarily expensive, but rarely used technique as you need to book a time with this lab waaaayyy in advance.)
Your question is merely an exercise in geometry! Should be a more comfortable field for pre-uni level chemists. It does highlight the different bond lengths, throwing out any pre-conceptions that all C-C bonds are equal, but the question, is essentially does an octagon fit in a cube. (If we boil it down to the absolute bare bones)
To take this question to a more advanced level you can then consider the fact that the carbons and H's are not infinitely small dots but occupy a significant space. If we take the radius of the carbon atom to be it's covalent radius, then for cubane we have 4 primitive cubic packed carbons. This has a packing efficiency of ~52%, so even if cyclooctane would fit in a cubane formed from infinitely small carbons, it won't fit in real cubane as the available space is actually much smaller than that of a cube of the same dimensions.
You could advance this even further by considering the actual conformation of cyclooctane. Cyclooctane is not an Octahedron! Check out the major conformers on the wiki page
http://en.wikipedia.org/wiki/CyclooctaneThis makes the geometry of the problem more difficult as the cyclooctane is no longer a regular polyhedron but can probably be modeled quite reasonably as a sphere of radius equal to the dimensions of the conformer of cyclooctane.
Yes the bond lengths are very similar to cyclobutane!