The maximum volume of the rectangular box is 8161.59 cm^3.
Let x, y and z be the length, width and height of the box respectively.
The volume of the box is V=xyz. We can find V in terms of x and y.
The surface area of the box is 1000 cm^2, so we have 2xy+2xz+2yz=1000.
Dividing both sides by 2 gives xy+xz+yz=500.
The total edge length of the box is 160 cm, so we have 2x+2y+2z=160.
Dividing both sides by 2 gives x+y+z=80. We can rewrite this as z=80-x-y.
Substituting this into xy+xz+yz=500, we get xy+x(80-x-y)+y(80-x-y)=500.
Expanding and simplifying, we get 2xy-2x^2-2yx+80y-80x-y^2=500.
Combining terms, we get -x^2-2x-y^2=200.
Rearranging, we get x^2+2x+y^2=400.
Completing the square, we get (x+1)^2+y^2=401. This means that x+1=√401=20.1 and y=√401-20.1=20.1. Substituting these values into V=xyz, we get V=xyz=20.120.120=8161.59 cm^3.
Therefore, the maximum volume of the rectangular box is 8161.59 cm^3.