data rate given as = 10 Mbps
Tt = 10 x 2 10 / 10 x 106 = 1024 / 106 = 1024 x 10-6 sec
Tp = 270 x 10 -3 sec
Capacity of a channel = 2*Tp* BW
= 2* 270 x 10 -3 * 10 x 106
= 5400 x 10 3 bit
No of packets sent = 5400 x 10 3 / 10 x 210
= 527.34 = 527 packet
you can send 527 packet
but In stop and wait we only sent 1 packet
therefore link utilization : 1/527 = 0.001897 = 0.18 %
just go with the concept