See I did like this. Let me know if I am in the correct direction.
The sequence number bits $m=8$
Protocol is Selective repeat, so $W_s=128$
Now since this protocol, unlike GBN, allows out of order packets, NAK will be generated only for corrupted packets.
The error rate for a data packet=1%.
So, the number of NAK generated(hence the number of retransmission needed)=$\lceil 0.01 \times 128 \rceil=2$
So, in total sender transmitted 130 frames.
Total data transmitted=$(130 \times 4000)+(2 \times 40)(NAK)=520080$
Overhead data sent=$(128 \times 40)+(2 \times 4000)(retransmissions)+(2 \times40)(NAK)=13200$
I am getting another answer :(