Are you familiar with stars and bars method?
So we have to find the number of 6-tuples $(x_1,...x_6)$ such that each $x_i$ is even number grater than $0$ and their sum is 100. So we can write each $x_i=2y_i$ and we have $$y_1+y_2+...+y_6 = 50$$
If we write $z_i= y_i-1$ we have $$z_1+z_2+...+z_6 = 44$$ and that we can write on $${44+5\choose 5}$$ ways.
Labeling the colors $A,B,C,D,E,F$ for conciseness, and using an ordered list of sets of colors to represent how the boxes were filled with balls,
consider the arrangement
$$ \{A, B, C\}, \{D\}, \{E\}, \{F\}. $$
You have counted this arrangement six times.
One way you counted it was by making the grouping
$ \{A, B, C\}, \{D\}, \{E\}, \{F\} $
and putting these sets of balls in the boxes in exactly the order they are listed here (i.e., the identity permutation).
Another way you counted it was by making the grouping
$ \{A, B, C\}, \{E\}, \{D\}, \{F\} $
and swapping the middle two sets before putting the sets of balls in the boxes.
The other four ways you counted this arrangement correspond to the other four
permutations of the last three subsets.
Dividing by $6$ won't help, because you counted each of the groupings such as
$ \{A, B\}, \{C, D\}, \{E\}, \{F\} $
exactly four times.
A correct count:
Each of the $\binom{6}{3,1,1,1} = 120$ groupings of the first kind can be distributed into boxes in $\binom 41 = 4$ different ways
(choose which box gets the set of three,
but keep the relative order of equal-sized sets the same).
Each of the $\binom{6}{2,2,1,1} = 180$ groupings of the second kind can be
distributed into boxes in $\binom 42 = 6$ different ways
(choose which boxes get sets of two, etc.).
So that's a total of
$$ 120 \cdot 4 + 180 \cdot 6 = 1560 $$
arrangements.
Note from the comments that $4!{6\brace4} = 24 \cdot 65 = 1560.$
Best Answer
First we assume that the balls are indistinguishable.
Line up the $5$ balls like this $$ B\qquad B \qquad B \qquad B \qquad B$$ We will choose $2$ from the $4$ gaps between $B$'s to put a separator into. Then all $B$ up to the first separator go into the first box, all $B$'s between the two separators go into the second box, and the rest go into the third.
There are exactly as many ways to insert separators as there are ways to distribute the balls between the boxes. So there are $\binom{5-1}{3-1}$ ways to do the job.
Remark: The idea generalizes. Please see the Wikipedia article on Stars and Bars.
Next we assume the balls are distinguishable. We use Inclusion/Exclusion. There are $3^5$ ways to distribute the balls, with no restriction.
There are $2^5$ patterns in which the first box is empty, and the same number with the second empty, and the same with the third empty.
However, if we subtract $3\cdot 2^5$ from $3^5$, we have subtracted too many times the patterns in which two of the boxes are empty. So we need to add back $3\cdot 1^5$, giving count $3^5-3\cdot 2^5+3\cdot 1^5$.