Question 1021148
I will assume that the iron bar has a square cross-section.
The diagonal of the square being 6cm ==> each side of the square is {{{6/sqrt(2)}}} cm.
==> the largest possible cylindrical rod would have a radius of {{{3/sqrt(2)}}} cm.

==> Volume of cylindrical rod is {{{V = Bh = pi*((3/sqrt(2))cm)^2*(200cm) = 900*pi (cm)^3}}}
But the volume of the iron bar is {{{V = LWH = (200cm)((6/sqrt(2))cm)^2 = 3600 (cm)^3}}}

Therefore the amount of waste is 

{{{3600 (cm)^3 - 900*pi (cm)^3 = 772.57 (cm)^3}}}, to two decimal places.