For $50\%$ utilization with Stop-and-wait,
$\large\dfrac{t_t}{t_t+2t_p}\geq \dfrac{1}{2},$
where, $t_t-$ Transmission time, $t_p-$ propagation delay. Here, $t_t = \frac{L}{B},$ where $L$ is the frame length in bits and $B$ is the bitrate of the channel.
$\large2t_t\geq t_t+2t_p$
$\large t_t\geq 2t_p$
$\dfrac{L}{B}\large\geq 2\times t_p$
$L\geq 2\large\times t_p\times B$
$\implies L=2\times 20\times 10^{-3}\times 4\times 10^{3}=160\text{ bits}$
So, answer is D.