Here, maximum segment size(1 MSS)=1460B.,so in one RTT 1460 B is transferred.To send 1000KB we need 1000*1024/1460 MSS= 702 MSS (appx).
Using slow start algorithm,
1 MSS | 2 MSS | 4 MSS | 8 MSS | 16 MSS | 32 MSS | 64 MSS | 128 MSS | 256 MSS |512 MSS | 1024MSS.So in total we need 10 RTT (10* 100 ms=1000 ms) to send 1000KB.
We know , throughput is the number of bytes sent per second.
So, to send 1000KB we need 1000 ms.So, in 1 sec, it will send 1000 Kb.
So,throughput =1000kbps=1 mbps(D)