In IEEE 802.3 Ethernet,
Utilization U= 100 * (F/R) / (F/R + T*C)
where , U= Percentage of Utilization
R = Transmission Rate
F = Number of bits in a frame
T = Slot Time
C = Number of Contention Intervals = (1-1/N)1-N - 1
N = Number stations
Frame size = (256 *8 ) bits = 2048 bits
Data rate is 10 Mbps
F/R = 2048 bits/10 Mbps = 204.8 µsec
Number of stations, N = 128
C = (1-1/N)1-N – 1
= { 1- (1/128) }-127 - 1
= (0.992) -127 - 1
= 2.7 - 1
= 1.7
U = (100 * 204.8) / ( 204.8 + 51.2* 1.7)
= 20480 / (204.8 + 87.07)
= 20480 / 291.84
= 70.17
Hence, channel utilization is 70%.
so answer is option A , 70%