Starting address $ = 190.100.0.0/16\space\rightarrow\space$ which means that the first $16$ bits are used for identifying network and next $16$ bits are available for host addresses. So the available addresses are from $190.100.0.0$ to $190.100.255.255$.
We have $3$ groups. So we can make use of the leftmost $2$ bits of the available address to identify a group. I am taking $00, 01$ and $10$ for these groups.
Group $1\space\rightarrow\space00\space\rightarrow64\space$Customers, $256\space$Addresses each.
For $256$ Addresses we need $8$ bits. So final $8$ bits are given for the addresses. For $64$ customers we can make use of the preceding $6$ bits. Hence this group would be,
ie; $190.100.0.0$ to $190.100.63.255$
Group $2\space\rightarrow\space01\space\rightarrow128\space$Customers, $128\space$Addresses each.
$7$ bits are needed for $128$ addresses and preceding $7$ bits are needed for $128$ customers.
So, this group is: $190.100.64.0$ to $190.100.127.255$.
Group $3\space\rightarrow\space10\space\rightarrow128\space$Customers, $64\space$Addresses each.
This group is identified with $10$ and after reserving $6$ bits for $64$ addresses, and $7$ bits for $128$ customers, we still have $1$ bit left. This goes partially unused, and here it can be either $0$ or $1$. I am taking $0$.
Therefore addresses in this group is : $190.100.128.0$ to $190.100.159.255$.
Hope this helps.
