In IPV4 header Identification number field is $16$bit, and an ID number is associated with each packet means you can assign unique ID number to $2^{16}$ packets and each packet contains $1024$ bytes of data.
So, You must transfer $2^{16}*1024B$ of data within $10$ seconds.
So, BW $\color{navy}{= \frac{2^{16}*1024*8\;bits}{10\; sec} = 53.68\;Mbps}$