Without loss of generality, let the radius of the circle be 2. The radii to the endpoints of the chord, along with the chord, form an isosceles triangle with vertex angle $ 120^{\circ} $. The area of the larger of the two regions is thus $ 2/3 $ that of the circle plus the area of the isosceles triangle, and the area of the smaller of the two regions is thus $ 1/3 $ that of the circle minus the area of the isosceles triangle. The requested ratio is therefore $ \displaystyle \frac{\frac{2}{3}\cdot4\pi+\sqrt{3}}{{\frac{1}{3}\cdot4\pi-\sqrt{3}}} =\frac{8\pi+3\sqrt{3}}{4\pi-3\sqrt{3}} $, so $ abcde\!f=8\cdot3\cdot3\cdot4\cdot3\cdot3=2592 $, and the requested remainder is $ \boxed{592} $.