The surface area of a cuboid is given by 2*(lw + wh + lh) where l, w and h are the length, width and height. The volume is given by l*w*h. For the volume to be maximum, all the dimensions of the cuboid should be equal in length. This can be proved but I do not think that is essential here, and we take it as an identity. So the solid we have is a cube.
The surface area of the cube would be 6*s^2, where s is the side.
6s^2 = A
=> s = sqrt[A / 6]
The volume is given by s^3 = [sqrt (A/6)]^3
=> (A/6)*sqrt (A/6)
The required maximum volume is (A/6)*sqrt (A/6).