Let $G = \langle 3 \rangle$. An arbitrary element of $G$ is of the form $3n$ for some integer $n$.
Define the map $\phi : G \to \mathbb Z_4$ by $\phi(3n) = [n]$, where $[n]$ denotes the equivalence class of $n$, modulo $4$.
First we note that $\phi$ is well-defined, because if $3n = 3m$ then $n = m$, so $[n] = [m]$.
Now we verify that $\phi$ is a homomorphism:
$$\phi(3n + 3m) = \phi(3(n+m)) = [n+m] = [n]+[m] = \phi(3n) + \phi(3m)$$
It's clear that $\phi$ is surjective. Now what is the kernel of $\phi$? It is the set of all $3n \in G$ such that $\phi(3n) = [n] = $. Since $[n] = $ holds if and only if $n$ is a multiple of $4$, the kernel of $\phi$ is precisely the set of multiples of $12$, in other words $\langle 12 \rangle$.
The first isomorphism theorem now allows us to conclude that $G / \langle 12 \rangle \cong \mathbb Z_4$.