Proceeding with the assumption I made in my comment: in the ones place we have
4 + 5 ≡ 3 (mod b)
9 ≡ 3 (mod b)
6 ≡ 0 (mod b)
or equivalently, 0 + bk = 6 or bk = 6 for some integer k. Then b must divide 6, so b is one of {1, 2, 3, 6}. But in base n, any given number can only be composed of the digits {0, 1, 2, ..., n - 1}, so it must that b = 6.
Just to confirm:
4₆ + 5₆ + (9)₆ = (6 + 3)₆ = 13₆
2₆ + 3₆ + 1₆ = (6)₆ = (6 + 0)₆ = 10₆
3₆ + 1₆ + 1₆ = 5₆
and so
324₆ + 135₆ = 503₆
as required.
