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