ok so
what we do is
how much greater=wasted space=cylindervolume-conevolume
volume of a cinlinder=hpir^2
volume of a cone=(1/3)hpir^2
we can alread sybsitute something
(3/3)hpir^2-(1/3)hpir^2=(2/3)hpir^2
so wasted space=(2/3)hpir^2
diameter=4
d=2r
d/2=r
4/2=2=r
and 6=height
so
2/3(6)(3.14)(2^2)
4(3.14)(4)
16(3.14)
50.24 in^3 wasted
