$\text{Network A sends an IP packet of 180 bytes of data + 20 bytes of TCP header}$
$\text{+ 20 bytes of IP header to B}$.
IP layer of $B$ now removes $20\text{ bytes}$ of IP header and has $200\text{ bytes}$ 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 $60\text{ bytes}$ of IP headers and $20\text{ bytes}$ 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 $180\text{ bytes}$ of application data are transferred from $A$ to $C$ and this causes $260\text{ bytes}$ to be transferred from $B$ to $C$.
Time to transfer $260\text{ bytes}$ from $\text{B-C}=\dfrac{260\times 8}{(512\times 1000)}$
$=\dfrac{65}{16000}=\dfrac{13}{3200}$.
So, data rate $=\dfrac{180\times 3200}{13}=44.3\text{ kBps}= 44.3 \times 8 = 354.46\text{ kbps}.$
Correct Answer: $B$