Do we consider the dissociation of water when dealing with ion concentrations of a solution?
It dissociates into 7 parts; copper and sulfate ions and five moles of water.

Water is also the solvent, so these don't count.

However the answer "two" is indeed not perfectly right, because the sulfate ion would (partially) react with water, to form 

$$\ce{SO4^2- + H2O &lt;=&gt; HSO4- + OH-}$$

A sulfate ion is so to say the opposite of sulfuric acid, which reacts

$$\ce{H2SO4 + 2 H2O -&gt; HSO4- + H3O+ + H2O &lt;=&gt; SO4^2- + 2 H3O+}$$

$\ce{HSO4-}$ is not strongly acidic anymore $(\mathrm{p}K_\mathrm{a} = 2),$ so the sulfate ion logically must react somewhat basic. (OK, $\mathrm{p}K_\mathrm{a} = 2$ is quite acidic, acetic acid has $\mathrm{p}K_\mathrm{a} = 4.8.)$

So the correct answer is slightly larger than two. But from the given possibilities in this multiple-choice test, two is by far the most reasonable.

(Update: The hydrated $\ce{Cu^2+}$ ion might well be acidic enough to overcompensate the basicity of the sulfate. In which case you'd still have more than two equivalents of ions, just it'd be a small amount of $\ce{H3O^+},$ not $\ce{OH^-}.$ And of course there is always the autoprotolysis of water.)