There are many ways this question can be interpreted. It depends if all the balls could be placed in each box or if only one ball goes in each box and there are leftovers. Also, are the boxes different from each other and are the balls different from each other?
For my answer, we will assume that each of the boxes is different and each of the balls are the same, so an equivalent question would be in how many ways can we place 6 identical balls into the 4 boxes. For example, one way would be for all 6 balls to be in the first box (6,0,0,0) which would be different from all 6 in the second box (0,6,0,0).
Written this way, we can see that permutations are involved for each of the possibilities and we will have to list all of the possibilities. These possibilities are all sets of 4 whole numbers that add to 6. There is probably a formula for this, but it can also be written out as these numbers are fairly small.
The possibilities and the permutation computation (Divide the factorial by the factorial of identical sets.) for each is:
(6,0,0,0) = 4!/3! = 4
(5,1,0,0) = 4!/2! = 12
(4,2,0,0) = 4!/2! = 12
(4,1,1,0) = 4!/2! = 12
(3,3,0,0) = 4!/(2!2!) = 6
(3,2,1,0) = 4! = 24
(3,1,1,1) = 4!/3! = 4
(2,2,2,0) = 4!/3! = 4
(2,2,1,1) = 4!/(2!2!) = 6
Finally add all the options: 84 ways