We need the maximum throughput, and for that we need to send as much data
as possible to fully utilize the bandwidth.
so, maximum packets that can be sent $=1 + 2a = 9$ (after calculation) for $100\%$ efficiency.
But we have a window size of 5 only, so we can send only $5$ packets at max.
Efficiency $=\dfrac{5}{9}$
Now, $\dfrac{A}{Q},$ Bandwidth of the channel $(BW) =\dfrac{L}{T_t}$
$=\dfrac{1000}{(50\times 10^{-6})}$
$= 20\times 10^{6} \text{bytes/sec}.$
So, max. throughput achievable $=\text{Efficiency$\times $BW}$
$=\dfrac{5}{9}\times 20\times 10^{6}= 11.11\times 10^{6} \text{bytes/sec}.$