Answer:
"115200 and 460800" is the correct approach.
Explanation:
Total Bytes = 98 Data Bytes + 2 extra label bytes for every 10ms
= 100 bytes (every 10m)
Rate of data for every second will be:
= [tex]\frac{100\times 8}{10\times 10^{-3}}[/tex]
= [tex]\frac{800}{0.01}[/tex]
= [tex]80000[/tex]
Consequently, the 115200 as well as 460800 baud rates seems to be appropriate.