Basically, Acknowledgement will be received by A will be 80 millisecond. Now,
A can send 128kb in 1 sec (bandwidth),
In 1 sec ----→ 128kb
1 sec ---→ (128kb/(32*8 bits)) packets ( As each packet of size 32*8 bits)
solving above, A can send 500 packet per second
So, In 80ms ---→ 500* 80*10^-3 = 40 packets
So, A can send 40 packets before it receives an acknowledgement.
So, Optimal window size for will be 40