Let's call the centers of the two circles O and P, and the points where they intersect Q and R. Then the area of the region labeled B is easily given in terms of the angles s=<QOR and t=<QPR by r2(s-sin s)+q2(t-sin t), since we can think of it as a union of two wedges (the positive terms) minus the corresponding triangles (the negative terms).
Now s is twice one angle of a triangle with sides r, r, and q, so from the law of cosines we have q2=r2+r2-2r*r*cos (s/2), or s=2*cos-1 (1-q2/2r2).
Now that we know what s is, t is just pi-s/2, so t=pi-cos-1 (1-q2/2r2).
Substituting s and t and simplifying, we get the following for the area of B in terms of q and r:
r2(2*cos-1 (1-q2/2r2)-sin 2*cos-1 (1-q2/2r2))+q2(pi-cos-1 (1-q2/2r2)-q/r * sqrt (1-q2/4r2))
We want this to be pi*r2/2, so we want q/r to be equal to z, where z satisfies:
pi/2=(2*cos-1 (1-z2/2)-sin 2*cos-1 (1-z2/2))+z2(pi-cos-1 (1-z2/2)-z*sqrt(1-z2/4))
which, assuming I haven't made any mistakes, should be easy to solve numerically, and probably not possible to solve analytically.