Packet $A$ sends an $IP$ packet of $180$ bytes of data $+ 20$ bytes of TCP header $+ 20$ $bytes$ of IP header to $B$.
$IP$ layer of $B$ now removes $20$ $bytes$ of $IP$ header and has $200$ bytes of data. So, it makes $3$ IP packets - $[80 + 20, 80 + 20 , 40 + 20]$ and sends to $C$ as the Ip packet size of $B$ is $100$. So, $C$ receives $260$ bytes of data which includes $60$ bytes of $IP$ headers and $20$ bytes of TCP header.
For data rate, we need to consider only the slowest part of the network as data will be getting accumulated at that sender (data rate till that slowest part, we need to add time if a faster part follows a slower part).
So, here $180$ bytes of application data are transferred from $A$ to $C$ and this causes $260$ $bytes$ to be transferred from $B$ to $C$.