What is the optical nature of the tris(ethylenediamine)cobalt(III) cation?
Ethylenediammine or $en$ is an symmetrical didentate ligand. You can remember a rule of thumb for this if you like:


  In the case of $3$ symmetrical didentate ligands, i.e. $[\ce{M(AA)3}]$, the compound exists in only one form, which is optically active


So the complex you've mentioned will be optically active, no doubt.

Also, as a reference, I'm quoting the wikipedia article at https://en.wikipedia.org/wiki/Tris(ethylenediamine)cobalt(III)_chloride#Stereochemistry


  The complex can be resolved into enantiomers that are described as $\Delta$ and $\Lambda$.


The only symmetry elements that can be found in $\ce{[Co(en)3]^3+}$ are the three $C_2$ axes of rotation, one of which is shown in this image (colored in red) on wikipedia: https://upload.wikimedia.org/wikipedia/commons/e/e5/Coen3lel3.png





As a side note, here's another rule of thumb:


  In the case of $3$ unsymmetrical didentate ligands, i.e. $[\ce{M(AB)3}]$, the compound exists in two forms, namely fac (stands for facial) and mer (stands for meridional), and both of these forms are optically active.




Important note: There's a minute error in this answer; and that's the distinction between optically activity and chirality: I used the word "optically active" while it should've been "chiral". It's bad I missed it, but I'd like you to read Ben Norris' comment on this below. I'm leaving the answer untouched, editing it rip the comment of its value. I encourage you to learn from my mistake.