jmc

Regions I, II, and III combine to form a sector of a circle whose central angle measures 90 degrees. Therefore, the area of this sector is $\frac{90}{360}\pi (\text{radius}{)}^2=\frac{1}{4}\pi (2{)}^2=\pi$ square centimeters. Also, regions I and II combine to form an isosceles right triangle whose area is $\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}(AB)(BC)=\frac{1}{2}(2)(2)=2$ square centimeters. Subtracting these two areas, we find that the area of region III is $\pi -2$ square centimeters. Since region II is congruent to region III, the combined area of region II and region III is $2(\pi -2)=2\pi -4\approx \boxed{2.3}$ square centimeters.