ThheBaIkanBeer

The answer is

Let the boxes be B1 and B2.

It it obvious, that under given condition, EITHER box B1 OR box B2 must be empty.

If box B1 is empty, then all 8 indistinguishable balls are in box B2: so, there is only one such distribution.

If box B2 is empty, then all 8 indistinguishable balls are in box B1: so, there is only one such distribution.

In all, there are 1 + 1 = 2 such distinguishable distributions.

p = mass of one ream of paper in kg
(2/7)p = mass of the empty box in kg

A = mass of empty box and 3 reams of paper
A = (2/7)p + 3p 
A = (2/7 + 3)p 
A = (2/7 + 21/7)p 
A = (23/7)p 
Note that 3 reams is 1/5 of the total 15 reams.

B = mass of empty box and 15 reams of paper
B = (2/7)p + 15p 
B = (2/7 + 15)p 
B = (2/7 + 105/7)p 
B = (107/7)p 

Mass A is 33.6 kg lighter compared to mass B.
A = B - 33.6
(23/7)p = (107/7)p - 33.6
7*(23/7)p = 7*( (107/7)p - 33.6 )
23p = 107p - 235.2
23p-107p = -235.2
-84p = -235.2
p = -235.2/(-84)
p = 2.8
Each ream of paper has a mass of 2.8 kg.

(2/7)*p = (2/7)*2.8 = 0.8 kg is the mass of the empty box.

Notes:
2.8 = 14/5
0.8 = 4/5