computer network closed

Question

On a TCP connection, current congestion window size is Congestion Window = 4 KB. The window size advertised by the
receiver is Advertise Window = 6 KB. The last byte sent by the sender is LastByteSent = 10240 and the last byte
acknowledged by the receiver is LastByteAcked = 8192. The current window size at the sender is
A) 2048 bytes
B) 4096 bytes
C) 6144 bytes
D) 8192 bytes

I am getting b as answer but answer given in gate book is A

Comments

sender window = min(congestion window,receiver's window) = 4KB, I think B

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Min(congestion window, advertised window) = sender window size

Here it is 4KB

Lstbytesend -lastByteack = 10240-8192 = 2048B

2048B = 2KB not acknowledged...

Current window size 4KB -2KB = 2KB