Answer : 33 bits
Given Bandwidth = $1Gbps$, MSL = $60 sec$.
Now, in $1$ sec, amount of data consumed = $1Gb$
Hence. in $60 sec$, amount of data consumed = $60Gb$ $\implies$ $(60Gb/8)$ $\implies$ $7.5GB$
Therefore, to prevent the sequence number space from wrapping around during the MSL, we need $7.5GB$ of data which can be represented with:-
$\lceil log(7.5 * 2^{30}) \rceil$ $\implies$ $\lceil log(7.5) + log(2^{30}) \rceil$ $\implies$ $\lceil 2.9+30 \rceil$ $\implies$ $33 bits$ for the sequence number field.
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The Answer will also be same if we use $1Gbps = 10^9bps$
Minimum no. of bits required in Sequence number field of the $\text{TCP}$ header to prevent the sequence number space from wrapping around during the $\text{MSL}$ is given by:-
$\lceil log(MSL*B_w) \rceil$ where $MSL$ (maximum segment lifetime) = $60 \; \text{seconds}$ [given]
and $B_w$ (Bandwidth in $Bps$) = $10^9bps = 10^9/8 Bps$ [given]
Thus,
Min. no. of bits required= $\lceil log(60* 10^9/8) \rceil$
$=\lceil log(60*10^4) + log(10^5/8) \rceil$ [Breaking $10^9$ so that it becomes easy to use the scientific calculator in exam]
$=\lceil19.19460+ 13.60964 \rceil$
$=\lceil 32.80424\rceil = 33bits$ for the sequence number field.