Option(B) is correct!
There are 3 ways to do this question.
1.
Tp = 400 ms, R.T.T = 2*400 ms = 800 ms
Tt = 40 ms
In one round trip, we only send 10 packets = 10*100 = 1000 Bytes = 8000 bits.
In 800 ms we send 8000 bits, throughput = In 1 second, how many bits can be sent?
1 ms = $\frac{8000}{800}$ = 10 bits, and in 1 second = 10 k bits
2.
$$\text{Throughput = efficiency*B.W}$$
efficiency = $\frac{n}{(1+2*a)}$ where $a=\frac{tp}{tt}$
efficiency = $\frac{10}{1+2*400/40}$ = $\frac{10}{21}$
Throughout = (10/21)*20Kbps = 10 Kbps (almost)
3.
R.T.T = 2*TP = 800 ms
Tt = 40 ms
in 1 R.T.T we can send maximum $\frac{800ms}{40ms}$ = 20 packets
But as mentioned in the question we are sending only 10 packets, it means efficiency is 50% (or) we are utilizing the bandwidth only 50%, so effective bandwidth or throughout will be 50% of total bandwidth i.e. 10 kbps.