If nothing is given , we can safely assume that the channel is full duplex.
What is Channel Capacity? → The amount of data that the channel can contain before the 1st bit reaches the receiver.
Channel capacity is also known as Bandwidth-Delay Product . For a half-duplex channel it is $B.W * T_p$.
$T_p = 2s$ and $B.W = 500*10^6 bps$ , so the Channel capacity for half-duplex channel is $10^9 \ bits$ , and that for a full duplex channel it will be $2*10^9 \ bits$ . Size of one packet = $10^7 \ bits$ , so the maximum number of packets that the channel can hold is $200$.
Now , squence number $>=$ $SWS+RWS >=200+1 = 201$ . So the minimum number of bits required = $log_2(201) = 8$ . Thus $8$ is the answer.