Answer: 864

Small Hint:

Recognize each $A$ as a square face with one corner cut off and each $B$ as that corner triangle

Big Hint:

A plane through six edge midpoints of a cube cuts a regular hexagon and divides the cube into two congruent parts

Solution:

Consider a $12\times12\times12$ cube and the plane through the midpoints of the six edges that join opposite groups of three vertices. In coordinates $0\leq x,y,z\leq12,$ this is the plane $x+y+z=18.$ Its cross-section is a regular hexagon with side $6\sqrt2,$ the same as the cut edge joining adjacent side midpoints.

On three faces of the cube, the plane leaves a square with a corner triangle removed, exactly piece $A.$ On the other three faces, it leaves the complementary right-isosceles corner triangle, exactly piece $B.$ Thus the pictured net is one of the two pieces into which this plane cuts the cube. Central symmetry interchanges the two pieces, so each has half the cube’s volume: $$V=\frac{12^3}{2}=864.$$