$255.255.255.224 = 11111111.11111111.11111111.11100000$
$192.168.1.97$
$192.168.1.80$
$192.168.1.135$
$192.168.1.110$
$192.168.1.67$
$192.168.1.155$
We need to do bitwise AND with subnet mask.
the last $5$ bits are going to be 0 when ANDED.
No need to waste time in finding binary.
Only focus on $1$st $3$ bits of binary.
(From left side, $1$st bit is $128$,next one is $64$,next one is $32$..it goes like that you know.)
$\bf{97}$: $0 + 64 + 32 +$ something so $1$st $3$ bits will contain $011$
$\bf{80}$: $0 + 64 + 0+$ something so $010$
$\bf{135}$: $128+0+0+$ something so $100$
$\bf{110}$: $0+64+32+$ something so $011$
$\bf{67}$: $0+64+0+$ something so $010$
$\bf{155}$: $128+0+0+$ something so $100$
So we got $011, 010, 100$
$3$ subnets.... subnet id are...
$192.168.1.96$
$192.168.1.64$
$192.168.1.128$
Correct Answer: $C$