Answer: B

Small Hint:

Equal parallel chords lie the same distance from the center, so each is $4$ inches from it

Big Hint:

Subtract the two congruent outer circular segments from the whole circle

Solution:

The equal chords are symmetrically $4$ inches from the center. For either chord, the half-angle at the center satisfies $\cos\theta=\frac{4}{8}=\frac{1}{2},$ so $\theta=60^\circ.$ Each outer segment is a $120^\circ$ sector minus the isosceles triangle: $$\begin{aligned} &\frac{120}{360}\pi(8)^2 -\frac12(8)^2\sin120^\circ\\ &\qquad=\frac{64\pi}{3}-16\sqrt3. \end{aligned}$$ The area between the chords is therefore $$\begin{aligned} &64\pi-2\left(\frac{64\pi}{3}-16\sqrt3\right)\\ &\qquad=\frac{64\pi}{3}+32\sqrt3\\ &\qquad=21\frac13\pi+32\sqrt3. \end{aligned}$$

Thus, the correct answer is B.