We can choose the 2 balls to go in the specified box in 10C2 = 45 ways.

For every way that can be done, there are 8 balls left to 
distribute among the other 3 boxes.

There are 3 decisions to make for each remaining ball. 
Those 3 decisions are whether:

1. to place it in the first box, or
2. to place it in the second box, or
3. to place it in the third box.

So we have 3 possible decisions for the 1st remaining ball.
And for each of those,
we have 3 possible decisions for the 2nd remaining ball.
And for each of those,
we have 3 possible decisions for the 3rd remaining ball.
And for each of those,
we have 3 possible decisions for the 4th remaining ball.
And for each of those,
we have 3 possible decisions for the 5th remaining ball.
And for each of those,
we have 3 possible decisions for the 6th remaining ball.
And  for each of those,
we have 3 possible decisions for the 7th remaining ball.
And  for each of those,
we have 3 possible decisions for the 8th remaining ball.

So the answer is 45*3*3*3*3*3*3*3*3 = 45*3^8

Note: This allows the cases when some of the boxes are empty,
and even some cases where all the remaining balls are all in 
one box.

Edwin