I calculate windows size with 10mbit bandwith and delay 5ms and the result was 6250 byte
No it isn't.
bandwidth = 10Mbits/s = 10*1024*1024/8 bytes/s = 1310720 bytes/s
delay = 5ms = 0.005s
product = 1310720*0.005 = 6553.6
However 6250 is far too small for a TCP send or receive buffer, and has been for many years. Linux probably agrees and enforces a higher minimum.