1760 yards.
Let the speed of the first ferry be u yards per minute, the speed of the second be v yards per minute, and the width of the river be L yards. Then the two ferries will meet after L/(u+v) minutes, at which point the first boat has traveled Lu/(u+v) yards from its station, which we are given to be 720 yards, so we find:
Lu/(u+v) = 720
Then each ferry continues across to the other station, waits at the station for 20 minutes, and returns to meet at the center. The first ferry leaves the opposite station at time L/u + 20; the second leaves at time L/v + 20. When will they meet (again)? They will meet at time T, where [T- (L/u + 20)] u + [T - (L/v + 20)] v = L, that is, when the total distance traveled by each ferry after they leave their rest stops is the width of the river. We can rewrite this as:
T (u+v) - 2L - 20(u+v) = L
T (u+v) = 3L + 20(u+v)
T = 3L/(u+v) + 20
An alternative way to produce T is as follows: Note that after meeting once already, the ferries must travel a total distance of 2L before they meet again, which they can accomplish in time 2L/(u+v). Add this to the time already spent reaching the first meeting point, L/(u+v), and add 20 minutes for the rest breaks they have, and you obtain the same result.
How far has the first ferry traveled from the second station to reach this point? The time it took to travel the distance is T - (L/u +20), or 3L/(u+v) - L/u, and its velocity is still u, so the distance is 3Lu/(u+v) - L, which we are given is 400 yards.
3Lu/(u+v) - L = 400
Use the relation we already know, that Lu/(u+v) = 720, and we obtain
3 * 720 - L = 400
L = 3 * 720 - 400 = 1760
If we want, we can continue to solve for the relative sizes of u and v. We know that Lu/(u+v) = 720, so (u+v)/u = 1760/720 = 22/9, so v/u = 13/9. Let us check our values for v = 130 yards per minute and u = 90 yards per minute. How long does it take the first boat to go 720 yards? It takes 720/90 = 8 minutes, in which time the second boat has gone 130*8 = 1040 yards, which is precisely the remaining distance to the opposite shore. The first boat continues to the opposite shore, which it reaches in a total amount of time 19+(5/9) minutes, and the second boat reaches its opposite shore at time 13+(7/13) minutes. Then the boats leave their respective shores when the clock reads 39+(5/9) minutes and 33+(7/13) minutes. How long does it take the first boat to travel 400 yards? Only 400/90 = 4+(4/9) minutes, bringing its clock up to 44 minutes even. The second boat then has 10+(6/13) minutes in the water, giving it time to travel 130 (10 + 6/13) = 1360 yards, which again is the exact distance it had to travel to meet the first boat. So the answer of 1760 yards passes the sanity check. Once you notice that the 20-minute wait has no ultimate bearing on the problem, then it becomes clear that only the ratio of the ferry speeds matters, making it plausible to solve a problem with three unknowns (L, u, and v) and only two pieces of information (the meeting points).