Answer Posted / jitendera kumar sinah

the given ip address is
172.16.210.0/22
scince it is class b address the it orignal mask should be 26 but given mask it 22 that is there is 6 bit of sunating
128+64+32+16+8+4=252
block size=256-252=4
now
210/4= 52
and remnder is 2(no matter)
that is
52*4=208
thus
subnet for the given host ip adress must be
172.16.208.0/22
thanks

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