1 bit length = (1/bandwidth)*propagation speed
= (1/4*106)*2.4*108
= 60 metre
now , to hold complete token of 24bit ,minimum length of the ring =24*1 bit length
=24*60=1440 m
now , one every station is providing 1 bit delay , so length covered by bit delay of all stations = n*60m ...............(1)
monitor is providing extra 15bit artificial delay ,i.e. 15*60 =900m .............................(2)
also , each station is 48m apart, hence physical length of ring = n*48...................(3)
now ,(1)+(2)+(3) shold be greater than or equal to 24 bit length i.e.1440
n*60+900m+ n*48 >=1440
so n>=5