We need only consider the triangular end of the barge to find the displacement. The triangle stands on its head with the base above the water line. Since the draft is 4m (the barge sinks 4m into the water) and the height of the triangle is 6m, the base is 2m above the water line. The displacement is the volume of water displaced by as much of the barge as is below the water line. So what we have is two similar triangles, a smaller one inside a bigger one. The big triangle is the cross-section of the barge, the small one has a base at the water line. We need to find the area of this small triangle. Because we have similar triangles we can write (base of cross-section)/(the base of the small triangle)=(height of the cross-section)/(height of the small triangle)=6/4=3/2. Base of the cross-section=12m. So we have: 12/(base of small triangle)=3/2. The base of the small triangle is 2/3*12=8m. Therefore the area of the small triangle is:
1/2(base)*(height)=1/2*8*4=16 sq m.
The volume of the triangular "prism" of water is 16*(length)=16*20=320 cubic m. Therefore the displacement is 320 m^3.