Assuming that by addresses we mean we can include the broadcast and the network ID in the block we allocate.
We have /22 means we have 10 bits in spare and 22 bits from the network ID part.
We need 100 subnets each of equal size, so SUBNET ID = ceil(100) = 7 bits and remaining 3 bits can be used for addresses allocated per subnet i.e. 2^3=8
So CIDR of first block = 220.127.116.11/29
Now we are left with 128-100 =28 Subnet ID each having 8 addresses. So 28×8=224 addresses are left with the ISP.