Dear User,
In order not to flood the buffers of B, the average number offrames entering and leaving B must be the same over a longinterval.

From A to B:

           Thepropagation delay time = 4000 * 5microsec = 20msec

The transmission time per frame = 1000/(100*10^3)=10msec

From B to C:

          Thepropagation delay time = 1000 * 5 micro sec = 5msec

The transmission time for frame = x=1000/R

The data rate between B and C =R

A can transmit three frames to B and then must wait for theacknowledgement of the first frame before transmitting additionalframes. The first frame takes 10 msec to transmit; the last bit ofthe first frame arrives at B 20 msec after it was transmitted andtherefore 30msec after the frame transmission began. It will takean additional 20msec for B’s ack to return to A. thus A cantransmit three frames in 50msec.

B can transmit one frame to C at a time. it takes 5+x msec forthe frame to be received at C and an additional 5 msec forC’s acknowledgement to return to A.thus,B transmit one frameevery 10 + x msec,or three frames every 30 + 3x msec.thus,

30 + 3x=50

X=6.66msec

R=1000/x=150kbps
I hope this will helps toyou