Answer
$Rate = (K_Fk_S)[A]^2[B]$
Work Step by Step
Slow, determining rate:
$r=r_S=k_S[I][B]$
Rate of formation of I
$r_{-F}=k_{-F}[I]=r_F$ (equilibrium)
$r_{F}=k_F[A]^2$
$K_F=k_F/k_{-F}\rightarrow [I]=K_F[A]^2$
Therefore:
$Rate = (K_Fk_S)[A]^2[B]$
