B)
Since Receiver window size is 6KB and N/W congestion window size is 4KB so sender has to send only 4KB window.
Sender window size= min(Wcongestion,Wadverstised)
=min(4KB,6KB)
=4KB
10240 B is last byte that is sent and last byte that is acknowledged is 8192.So free space in window will be 2048 B but question is just asking you Ws which will be 4KB.