**Answer: 7 bits**

For Go Back to N:

- 'N' buffers are maintained on sender side and
- 1 buffer is maintained on receiver side

So, we need unique N + 1 sequence numbers

So the bits required would be Ceil $log_{2}(N + 1)$ bits to represent these numbers.

Now let us find what the N should be.

- Since the propagation delay is 6 μsec/km, which means for 1 Km it takes 6 μsec to travel from one end to another
- So, for 3000 Km it would take 3000*6 μsec to travel from one end to another end, which is
** **18000 μsecs of total propagation delay.
- So
**RTT would be $2*18000$**

So, to achieve full capacity, sender must keep on transmitting for this time.

T1 trunk has a 1.544 Mbps transmission speed and is full duplex.

This means,

- In 1 sec, $1.544 * 10^{6}$ bits can be loaded on link
- Or $1,000,000$
**μ**sec, $1.544 * 10^{6}$ bits can be loaded on link
- Or 1 μsec, $\frac{1.544 * 10^{6}}{1000000}$ bits can be loaded on link
- So, In $2*18,000$ μsec, $\frac{1.544 * 10^{6} * 2*18000}{1000000}$ bits can be loaded on link
- Which comes to $55,584$ bits or $6948$ bytes

Frame size is given as 64 bytes. So $\frac{6948}{64}$ packets or 108 packets can be sent

So, our N is 108 packets, so $N + 1$ is $109$ packets

Bits needed for Ceil $log_{2}109$ which is *7 *bits