Solution

\(\)Key Points

 Given data,

Band width = B = 100 Mbps =10⁸ bps

Distance = D = 1 koilometer

Frame length = L = 1250 bytes

Speed = S =?

 Important Points

A minimum frame length for CSMA/CD protocol will  be T_x=2 T_p

Here T_x = transmission  delay = Frame length / Bandwidth = L/B

T_p = propagation delay =  Distance / speed of signal = D/S

Calculation:

A minimum frame length for CSMA/CD protocol will  be Tx= 2 Tp

\({L \over B } = 2{T_P} \\{L \over B } = 2{D \over S} \)

\({{1250 \times8} \over {100 \times 10^8}} ={{ 2 \times {10^3}} \over S} \\ S= 2 \times { ({2 \times 10^3 \times 100 \times 10^8 }) \over (1250 \times 8)} \\S= 2 \times 10 \times 10^3 km/sec \\ S= 20000\)

Hence the correct answer is 20000.