Let $X=S^1 \vee S^1$ and so that $\pi_1(X)=F\{a,b\}$, the free group on two generators. Let $\varphi:\pi_1(X) \rightarrow \mathbb{Z}/3$ be the homomorphism induced by $\varphi(a)=1$ and $\varphi(b)=0$. Draw the associated cover of the kernel of $\varphi$.
I think that the kernel should be the free group $\langle a^{3},b \rangle$.
I am not entirely sure what the question is asking.
There is an error in your description of the kernel, but as I think you are aware, it contains $a^3$ and $b$. Construct a covering space by drawing a triangle and labelling the sides "$a$," then attach a circle to each vertex and label it "$b$." This gives a degree $3$ covering map, whose image at the level of fundamental groups contains $a^3$ and $b$ and is indeed the kernel of the given homomorphism. However you should note, for example, that $aba^{-1}$ is in the kernel; in fact the kernel is a free group of rank $4$.
Let me know if you have any questions about the constructions.