You can do this by inclusion/exclusion. There are $\binom r0r^n$ ways of putting the $n$ balls into the $r$ boxes. From this we have to subtract the $\binom r1(r-1)^n$ ways of putting the $n$ balls into just $r-1$ of the boxes. To this we have to add the $\binom r2(r-2)^n$ ways of putting the $n$ balls into just $r-2$ of the boxes, and so on, so
$$a_n=\sum_{k=0}^r\binom rk(-1)^{r-k}k^n=\left.\left(q\frac{\partial}{\partial q}\right)^n(q-1)^r\right|_{q=1}\;.$$
Note: The formulation of this question has to be considered carefully in order to find the correct approach.
The situation describes $N$ indistinguishable balls which are to be distributed in $m$ distinguishable boxes.
Looking at a simple example as suggested by @MickA is often a good starting point. Let's have a look at an example with $N=2$ indistinguishable balls and $m=2$ distinguishable boxes. So, after distributing the $2$ balls into the $2$ boxes, we can see one of three possible outcomes:
\begin{array}{rcl}
\text{Box $1$} &|& \text{Box $2$} \\
\hline
\bullet\;\bullet &|& \\
\bullet &|& \bullet \\
&|&\bullet\; \bullet
\end{array}
We observe, if the question would have asked for something like
The Ansatz would be according to the first approach
\begin{align*}
\binom{N+m-1}{N}\tag{1}
\end{align*}
which respects the fact that the balls are indistinquishable and the boxes are distinguishable.
But this is not the question! The crucial text is:
... given that the balls have equal chances of arriving at any box.
Therefore we have to count the number of possible arrivals
\begin{array}{rclcc}
\text{Box $1$} &|& \text{Box $2$} &|& \text{Nr of possibilities}\\
&|& &|& \text{to reach this configuration}\\
\hline
\bullet\;\bullet &|& &|& 1\\
\bullet &|& \bullet &|& 2\\
&|&\bullet\; \bullet&|& 1
\end{array}
We see, counting the number of possibilities or equivalently calculating the corresponding probabilities
\begin{array}{rclcc}
\text{Box $1$} &|& \text{Box $2$} &|& \text{probability}\\
&|& &|& \text{to reach this configuration}\\
\hline
\bullet\;\bullet &|& &|& 0.25\\
\bullet &|& \bullet&|& 0.50\\
&|&\bullet\; \bullet&|& 0.25
\end{array}
is a different situation than (1) leading to the second approach
$$m^N$$
Result:
Therefore, we finally get the probability:
$$\left(\frac{m-1}{m}\right)^N$$
Best Answer
We count the number of ways of putting the balls into boxes by inclusion exclusion. There are $3^6$ ways of distributing the balls among the boxes, but we do not admit the ways of distributing the balls among the boxes where one box is empty, is ${3\choose 1}2^6$. Then we must add back in the times when two boxes are empty ${3\choose 2}1^6$. So, we get:
$$ 3^6-3\cdot 2^6+3\cdot 1 $$
Then we just need to calculate the number of ways that each box can end up containing two balls.
If we think of a permutation of the $6$ balls, and interpret the first two elements as going in the first box, the next two going in the second box, and the third two going in the third box, then we see that there are $6!$ ways of placing the balls evenly into the three boxes where the balls are ordered in the boxes. Then we have to divide by $2^3$ because the ordering of elements in their distinct groupings does not matter.
So, the final probability is:
\begin{equation} \frac{6!/2^3}{3^6-3\cdot 2^6+3\cdot 1}\approx 0.16667 \end{equation}