Option (A) 1007ms
propogation delay from Source computer (S) to R1 = (Distance) / (Link Speed) = 10^5/10^8 = 1ms
Total prorogation delay to travel from S to D = 4*1 ms = 4ms ( since four links are in between L1, L2, L3 and L4)
transmission delay for 1 packet = (Number of Bits in one packet) / Bandwidth = (1000/10^6) = 1ms.
total transmission delay for 1 packet = 4 * 1ms = 4ms ( since four links are in between L1, L2, L3 and L4)
The first packet will take 8ms to reach Destination Computer (D). While first packet was reaching D, other packets must have been processing in parallel. So D will receive remaining packets 1 packet per 1 ms from R3. So remaining 999 packets will take 999 ms. And total time will be 999 + 8 = 1007 ms