Since it will be a natural join among them, therefore on the basis of common attribute.(but ,since the is a patent child relation between the common attribute and parent is a primary key therefore this will be an interesting case)
So first let's take r1 and r2
Since, r2 have C as primary key therefore r1 cross r2 will have 1000 elements. Since if parent child relation exist therefore all the child's table rows of them will be taken.(your can check it via manually it's interesting)
Therefore now (ABCDE) cross r3
Similarly, 1000 is the answer.