Refer to the exhibit. What is the most efficient summarization that R1 can use to advertise its networks to R2?

The top two (4.128/25 and 4.0/25) aggregate into 4.0/25, so now we have four /24 networks. if they are aligned right, we can merge these into a /22. If not we’d need to merge them into a /21. The other three subnets are 5.0/24, 6.0/24, 7.0/24.
That means we start at 4.0 and go up to 5.255, i.e. just below 8.0. So this fits neatly into a /22 as these go up in steps of 4 in the 3rd octet. Yes, we can summarise with a /22 that starts at 172.1.4.0, answer D.