There is no need to use the pie. The following approach can help
consider each can as box of 4x4x12.
Now lets consider the following cases
1) the maximum cans that can placed considering that the cans occup full volume is
14x16x20
--------- = 23.3 so options D and E ruled out
4x4x12
2) now try option c if it fits
the surface area of carton is 14*16
the surface area of can 4*4 = 16
so total cans on surface area = 14(a)
we are still left with 20 - 12 = 8 units in height
placing can laterally we can have =14*16/4*12 =4.66(b)
still we have 4 units of space left which means total cans can be a+ 2*b ~ 22 ~20 cans
So my answer would be ©
What is the OA