Solution:

Let $O$ be the shared center. The rotation carries $A$ to $E,$ so $OA = OE$ and $\angle AOE = 20^\circ.$ That makes triangle $OAE$ isosceles, with base angles $\angle OAE = \frac{180^\circ - 20^\circ}{2} = 80^\circ.$ The diagonal $AC$ splits the right angle at $A,$ so $\angle OAB = 45^\circ.$ Subtracting, $\angle EAB = 80^\circ - 45^\circ = 35^\circ.$ Thus, B is the correct answer.