I am having difficulty trying to understand this question. All I know is that $3\mathbb{Z}$ and $\mathbb{Z}_6$ are both groups, that is:
$$3\mathbb{Z}=\{\dotsc, -6, -3, 0, 3, 6 \dotsc\}$$
$$\mathbb{Z}_6= \{0, 1, 2, 3, 4, 5\}$$
I took $\mathbb{Z}_6$ to be the set, and taking elements of $\mathbb{Z}_6$ to be represented by $m$. On the other hand, represented the elements of $3\mathbb{Z}$ with $n$. Hence, if I were to take $m=0$ then $n=0$ and applying the action:
$$(3(0), (0))=(3(0) + 0) \bmod 6 = 0$$
I repeated the procedure for:
$m=1$, $n=3$ giving $4$
$m=2$, $n=6$ giving $2$
$m=3$, $n=9$ giving $0$
And for some reason, I ended up with only: $\{0, 2, 4\}$ which I assumed to be my orbits. I am not sure whether I've applied the action correctly, for most of the work I've only made assumptions. In fact I would really need your help for enlightenment. Thanks.
To compute the orbits of $m\in\mathbb{Z}_6$, you need to find the set $\{3n+m\bmod6:n\in\mathbb{Z}\}$, as this is the set of all elements of $\mathbb{Z}_6$ obtained by acting on $m$ by some element $3n$ of $3\mathbb{Z}$.
Thus you need to compute $3n+m\bmod{6}$ for all $n\in\mathbb{Z}$ - note in particular that there is no relationship between $n$ and $m$.
You need to repeat this procedure for each element of $\mathbb{Z}_6$, although remember that orbits only intersect if they are equal. For example, you will compute that $3$ is in the orbit of $0$ (because $3=3\times 1+0\bmod6$), so the orbit of $3$ is equal to the orbit of $0$, and you don't need to calculate the orbit of $3$ separately.
If you know it, you could make things more efficient by using the Orbit-Stabiliser theorem to prove that the size of every orbit is $2$, which allows you to work out the number of orbits without having to compute the orbits themselves.