Official solution
Considering the three sub-cases for the remaining balls (0,5), (1,4), (2,3) split among the other two boxes:
\[\text{Favourable} = \frac{10!}{2!3!0!5!}4!+\frac{10!}{2!3!1!4!}4!+\frac{10!}{(2!)^2(3!)^2 2!3!}4! = 2^5\times17\times945\]
\[P = \frac{2^5\times17\times945}{4^{10}} = \frac{17\times945}{2^{15}}\]
This does not match any listed option cleanly, so the question was marked as Bonus in the source.