class C default NetWork mask : 255.255.255.0 ..But subnet mask is 255.255.255.224
224 = 11100000 = 128 + 64 + 32
so, 3 1's have been borrowed from host bits which will now become subnet bit.
for subnets there are 8 subnets ( 23 = 8 ) possible like 000, 001, 010 , 011 and so on ..... the first subnet is 000, second subnet 001, third 010 and fourth 011, fifth 100 , sixth 101 , seventh 110 , eight subnet 111 .
Hence subnet no of 6th subnet = 101 ,
The address with all 1s ( 111 11111 ) in host part is broadcast address and can't be assigned to a host. So the maximum possible last octal in a host IP is 111 11110 ,
here it asked 6th subnet so last octet of last host will be 101 11110 .
The decimal value of last octet of last host of sixth subnet is 101 11110 = 128+32+16+8+4+2 =190.