I assume that the isosceles triangle ACE and parallelogram BDFG are in parallel planes a distance 6 units apart, which appears to be the minimum length of AB (if AB were perpendicular to CD, which it may not be). I also assume that in BDFG, BD=FG=8 units, and BF=DG=5 units. Angle ABF is a right angle.
The height of isosceles triangle ACE = 4 units because, the perpendicular from E on to AC=√(5²-(6/2)²)=4. And the distance between BD and FG is also 4, because the perpendicular from B on to FG=√(5²-3²)=4. So ACE fits snugly into BDFG. And we already know that CE=DG=5 units so ECDG is also a parallelogram.
If we create a triangle AEF congruent to ACE we can make a parallelogram ACEF so creating two parallelograms for the top and base.
We know that the area of a triangle or parallelogram is determined by its base length and height and is unaffected by the skewness of its sides. That being the case, we can take AB=6 units, because lateral displacement of the top and bottom surfaces is not going to affect the volume. It also means we can replace the top and base by rectangles of the same area. We know the width is 4 units for top and bottom so the area of the base is 8×4=32 sq units and the top is 6×4=24 sq units. So what we effectively have is a cuboid and a triangular prism and we can easily find the total volume. The dimensions of the cuboid are 6×6×4=144 cubic units. The dimensions of the prism are 2×6×4, where 2 is the difference between the lengths of the top and bottom. We have to halve this product because of the triangular shape, so we have 24 cubic units. Total so far is 144+24=168 cubic units.
Now we have to subtract half of the smaller cuboid, because we added an extra piece created by the congruent isosceles triangle. The smaller cuboid has a volume of 24 cubic units, so half is 12 cubic units and the final figure for the volume of the gemstone is 168-12=156 cubic units (I think!).