The present best solution is using the concept of efficiency. I have tried solving the problem without using efficiency.
The time required to transmit 1000 bytes $= \frac{1000 \ bytes}{10 \times 10^6 \ bps} = \frac{8 \times 10^3 \ bits}{10^7 \ bps} = 8 \times 10^{-4} \ s = 800 \ \mu s $
$ \therefore \ $ The total time required to transmit a 1000 byte data $= (800 + 80) \ \mu s = 880 \ \mu s$
Now, in $ 880 \ \mu s $, we can send $8000$ bits.
$ \implies $ in $ 1 \ s $, we can send $ \frac{8000}{880 \times 10^{-6}} $ bits $ = \frac{8000}{880} $ Mb $ = \frac{100}{11} $ Mb