**Data:**

L = size of frame

BW → bandwidth = 20 × 10^{6} b/s

T_{p} → propagation time = 40 μs = 40 × 10^{-6} s

T_{t} → transmission time

**Formula:**

For CSMA/CD,

T_{t} ≥ 2 × T_{p}

\({{\rm{T}}_{\rm{t}}} = \frac{{\rm{L}}}{{{\rm{BW}}}}\)

**Calculation:**

\(\frac{{\rm{L}}}{{{\rm{BW}}}} \ge 2 \times 40 \times {10^{ - 6}}\)

L ≥ 2 × 40 × 10^{-6} × 20 × 10^{6}

L ≥ 1600 bits

L ≥ 200 bytes

L_{min} = 200 bytes

Therefore, the minimum size of a frame in the network is 200 bytes.