The smallest solution by area is the $4\times 2$ rectangle. The smallest solution with respect to longest side is the $3\times 3$ rectangle.
(1) The $3\times 3$ rectangle works:
N e N
R B R
e B e
(where N=kNight, R=Rook, B=Bishop, e=empty)
(2) The $4\times 2$ rectangle works:
N R R N
e B B e
(3) No $1\times n$ rectangle can work
Proof: Bishops and knights do not attack any other squares on such a rectangle; rooks only attack two squares. But altogether we need $2\cdot6=12$ attacks.
(4) The $3\times2$ rectangle cannot work
Proof: Consider the two middle squares. No knight can attack them; every bishop and every rook can attack at most one one of them. For having each of them attacked twice, the rooks and bishops must occupy the outer four squares, and the knights must be on the middle squares. Impossible.