Correct Answer - Option 3 : 64 bytes
Data:
L = size of frame
BW → bandwidth = 10 Mbps = 10 × 106 b/s
Tp → propagation time = 25.6 μs = 25.6 × 10-6 s
Tt → transmission time
Formula:
For CSMA/CD,
Tt ≥ 2 × Tp
\({{\rm{T}}_{\rm{t}}} = \frac{{\rm{L}}}{{{\rm{BW}}}}\)
Calculation:
\(\frac{{\rm{L}}}{{{\rm{BW}}}} ≥ 2 × T_p\)
L ≥ 2× Tp × BW
L ≥ 2 × 25.6 × 10-6 × 10 × 106
L ≥ 512 bits
L ≥ 64 × 8 bits // 8 Bit = 1 byte
Lmin = 64 bytes
Therefore, the minimum size of a frame in the network is 64 bytes.