In $GBN$ protocol , if we are using m bits of sequence numbers then the maximum SWS can be 2$^{m}-1$ and by default we start sequence numbers from $0$ .
In this question we have sequence numbers from 0 to 31 and the max $SWS$ will be of size $31$ from 0 to 30 and RWS is $1$ always.
Now it's given that the timer goes off for the frame containing sequence no 8.We need to understand the mechanism of the timer to solve this.In case of independent or cumulative acks the timer is started at the receivers end once the first packet is recieved.For independent acks it's very small just to ekaspse the processing and the queueing delays in the buffer while for cumulative we need an optimal to get the entire window accepted.The frame with 0 sequence number is received and the timer is started by shifting the RWS …
$0-->1-->2--....-->7--->8$ received and ack timer goes off
when the timer goes off for frame with sequence number 8 then it means we have already received that frame and the ack 9 is travelling towards the sender .
So the receiver will retransmit the frames with sequence numbers 9 to 30 if no upcoming frames are available,otherwise it will send to its max SWS capacity
SWS:Sender Window Size and RWS:Receiver Window Size
Correct answer will be option $C$.
