In this question, they are saying that 50 bits are globally allocated for unique identity field,
with 50 bits in the unique identity field, we can distinguish 2^ 50 B data uniquely
now total Byte spend in 1 sec is-
let assume at worst case all host are there so the total number of host =( 2^32) and each consuming 1000 B (for simplicity takes 1024) so
total byte consumed = (2^ 32) * (2^10) = 2^ 42 B
2^ 42 B is consumed in 1 sec, then 2^ 50 B is consumed in 256 sec (you please do the maths).