my public IP is
Why RIP packet contains such information?
I don't understand the IP notation and what it means.
IP Address: 18.104.22.168, Metric: 1
Address Family: IP (2)
IP Address: 22.214.171.124 (126.96.36.199)
RIP is Routing Information Protocol, and is a very basic (and early) protocol for announcing routes available on a given interface. It's a broadcast protocol. For RIP packets that do NOT contain a Netmask, like yours, it presumes the use of the old and very deprecated classful designation of Subnets. In your case, it is announcing a route for 188.8.131.52/24, the old, deprecated, Class C.
Specifically, RFC1058 section 3.2 defines how these no-subnet announcements work:
[Emphasis is mine]
RIPv1 had no field for subnet-mask. RIPv2 has one, but still allows absent subnet-mask.
In your case where network 89/8 was in the old Class A range, this suggests that your RIP packet is actually a v2 packet. V2 speaks CIDR, it's just optional, so the HOSTS getting the RIP announcement would evaluate it versus whatever subnet-mask they have on their own network configuration.
That address is out-of-range, so it would be seen as an announcement that the specified network, wherever it is, can be reached at the IP address of the source. As the announced address ends in Zero, and no subnet-mask was given, a host may assume that the netmask is /24.