Heats of combustion and stability of rings
Compounds A-D all have the same molecular formula, $\ce{C6H12}$.  We can burn each compound and measure the heat given off (heat of combustion). Since they are isomers, they will each burn according to the same equation

$$\ce{C6H12 + 9O2 -&gt; 6CO2 + 6H2O + heat}$$

Any differences in the heat given off can be used to say that a compound is more stable (it had a lower energy to begin with, so less heat is given off) or less stable (it had a higher energy to begin with, so more heat is given off).

This link provides the heats of combustion for some useful model compounds.  Look at the last column ("Total Strain") in Table I, it shows that cyclopropane is slightly more strained than cyclobutane, while cyclopentane and cyclohexane are both much less strained.  The strain energy (SE) in cyclopropane will not change appreciably when we add a propyl group to the ring, nor will the SE in cyclobutane change appreciably when we add an ethyl group to the ring.

Therefore, since cyclopropane has the most ring strain and since propyl, ethyl and methyl groups don't contain any SE, the correct answer to the question is "A".