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+2 votes

In LAN 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 = (512 *8 ) bits = 4096 bits

Data rate is 10 Mbps

F/R = 4096 bits/10 Mbps = 409.6 µsec

Number of stations, N = 500

But here C=No of Contentions given is= 1.7

U = (100 * 409.6) / ( 409.6 + 51.2* 1.7)

= 40960 / (409.6 + 87.04)

= 40960 / 496.64

= 82.4

Hence, channel utilization is 82%.

So Option B is the answer

for references: https://gateoverflow.in/96567/ethernet

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