For selective repeat: size of window( W ) = 2k-1
(Utilization or Efficiency)selective repeat = $W \div ( 1 + RTT / TD)$
Since utilization is given 100%,
TD = 8 * 103 / 128 * 103
= 8 / 128 sec
= 0.0625 sec
RTT = 2 * PD
= 2 * 150 ms = 300 * 10-3 = 0.3 sec
1 = W / (1 + 0.3 / 0.0625)
W = 5.8 or,
2k-1 = 5.8
k = 4 only this value satisfies.
Thus the min no. of sequence bits required for 100% utilization is 4.