In slow start phase, sender window size increases by 1 on receiving an acknowledgement from the reciever each time. So in 1 RTT, the sender window size gets doubled its previous value in slow start phase.

Now initial sender window size= 1MSS

receiver divides the resulting acknowledgment into M

This means that in place of sending 1 ACK for a segment, the reciever sends M ACKS for a single segment.

According to the slow start alogrithm, as long as congestion window size is less than or equal to the threshold, each time the sender recieves an ACK,it increases the congestion window size by 1. So when it recieves M acks for each segment, the congestion window size increases by M*number of segments the sender had sent, after each RTT.

inital congestion window size(cwnd)=1

After recieveing M acks, cwnd becomes= M+1 after 1RTT

now for these M+1 segments, sender recieves M(M+1) acks.

So cwnd after 2RTT= M+1+M(M+1)= M^{2}+2M+1=(M+1)^{2}

Similarly after 3RTT, cwnd= (M+1)^{2}+M(M+1)^{2}= (M+1)^{2}(M+1)= (M+1)^{3}

So after n RTTs, the sender window size will be (M+1)^{n}.

Hence (b) is the answer.