Since TCP is a byte stream protocol, every byte is numbered using a sequence number.

Every packet can loop around the internet for some period of time, which is the life time of the packet. Therefore, to ensure that there are no conflicting sequence numbers, we have to find out the Wrap Around Time of the sequence numbers after which the number is used again.

Now this Wrap Around Time $T_{WAT}$ should be greater than LifeTime of the packet $T_{LT}$, i.e $T_{WAT} > T_{LT}$

We have $32$ bits for sequence number field in the TCP header.

Given Bandwidth = $10^9$ bps(**bits** per second)

$\therefore 10^9$ **bits** can be sent out in $1$ second

$\therefore 2^{32} * 8 $ **bits** can be sent out in $x$ seconds ($\because$ TCP is byte streamed)

$\therefore x = \frac{2^{32} * 8}{10^9} = 34.359 $ seconds

Since they have asked the closest integer, answer is $34$

P.S Initially I was convinced that answer should be $34$, however, if we consider the statement "The minimum time (in seconds, rounded to the closet integer) before this sequence number can be used again", then I might incline towards $35$, although official key says $34$.