A $3000\ km$ long trunk operates at $1.536\ mbps$ and is used to transmit $64$ bytes frames and uses sliding window protocol. The propagation speed is $6\mu sec/km$

The minimum number of bits required in the sequence number field of the packet is

$T_t=333.33\mu sec$

$T_p=18000\mu sec$

Now,

$a=\frac{T_p}{T_t}$

$a=\frac{18000}{333.33}$

$a=54$

For sliding window protocol, sender window size required for maximum efficiency is:

$W=1+2a$

$W=1+2(54)$

$W=109$

So, sequence number field must be able to represent at least $109$ different sequence numbers.

Using $k$ bits, number of sequence numbers is:

$2^k$

Now check:

$2^6=64$

$64<109$, so $6$ bits are not sufficient.

$2^7=128$

$128\ge 109$, so $7$ bits are sufficient.