Efficiency of S&W protocol=$\frac{1}{1+2a}$

where a=$\frac{T_{p}}{T_{t}}$

propagation delay $T_{p}$=20 ms

Bandwidth of the channel=4Kbps=$4*10^3$ bits per second

transmission delay=$\frac{Length}{Bandwidth}$=$\frac{Length}{4*10^3}$

$\frac{1}{1+2a}\geq 0.5$

=$\frac{1}{1+\frac{2*20*10^{-3}*4*10^3}{L}} \geq 0.5$

On Solving, we get $L\geq 160$ bits

Frame size should be greater than 160 bits