a)Throughput for file transfer $= 500\text{kbps}$
b) Time taken for transfer the file to host $\text B=\frac{(4\times {10}^6)\times 8}{(500\times{10}^3} = 64\; \text{seconds}$
c)For R2, Throughput = $100\text{kbps}$, Time taken for transfer$ = \;\frac{4\times{10}^6\times 8}{100\times{10}^3}= 320\; \text{seconds}$