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2 Answers

Best answer
9 votes
9 votes
2 * propagation delay  + transmission time for jamming signal <= transmission time for frame
64 μs + 48/100 μs <= transmission time for frame
64.48 μs <= l/100 μs
l >= 6448 bits >= 806 bytes.
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6 votes
6 votes

L>= 2* Tp *B

L-> Length of frame

Tp -> Propagation time = RTT/2 = 64 * 10- 6/2 = 32 * 10-6

B -> Bandwidth = 10bps

So, L>= 2 * 32 * 10-6 * 10

     L>= 6400 b

Converting in Bytes 6400/8= 800 B 

So option 2

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