There are only Two cases a) and b) in which we get $2^{nd},3^{rd},6^{th},\text{and},\;7^{th}$ bits as $1$
1.
2.
In case 1) we are not reaching at final state with $8$ bits
From Case 2)
The number of $8$ bit strings in S that are accepted by M =2
1. 01110111
2.01110110
Hence,(c)2 is the correct choice.