@Randy Dandy,
Follow Poyla's rules
What do you know?
Hint---radius is half the diameter
1) You know the outside radius (r0) and height (h) of the cylinder,
So you know the outside volume of the cylinder ( V0=pi*ro^2*h)
2) You know the inside radius (ri) and height (h) of the cylinder,
So you know the inside volume of the cylinder ( Vi=pi*ri^2*h)
3) You know the cylinder is hollow (AKA annulus) --so the volume of the hollow cylinder (Vc) is the difference between the volume of the outside less the volume of the inside.
Vc=Vo-Vi
4) Units, Units, Units---The radius and height were given in inches--so if I use those measurements the Volumes calculated in steps 1,2 & 3 are in cubic inches (in^3)--- if I want to convert these volumes to cubic feet (ft^3)--I have to convert.
As there are 1728 in^3 in 1 ft^3 (1728=12^2). To get the volumes from in^2 to ft^3 divide by 1728.
5) The density of iron (not grade) is 441 lb per ft^3
To determine the weight of the hollow cylinder lb you multiply the volume of the hollow (annulus) cylinder in ft^3 by the density in lb/ft^3.
6) Check your answer--a 6 by 12 inch inch cylinder is smaller than a breadbox--consequently even though it is iron I should be able to pick it up, so it is probably less than 100 pounds. Now this annulus is hollow, so it should be significantly less than that.
Rap