The bandwidth of this channel is 36,000 bits per second.
How to solve
In a stop-and-wait protocol, the bandwidth efficiency can be calculated using the formula for efficiency, which is the ratio of useful data transmission time to the total time for one frame's round trip. Given an efficiency of 25%, the useful transmission time is 25% of the total time.
The total time for a frame's round trip includes two times the propagation delay:
2 x propagation delay=2 x 50 msec=100 msec.
Therefore, the useful transmission time is 25% of 100 milliseconds, which is 25 milliseconds.
Hence, the bandwidth of this channel is 36,000 bits per second.