Answer:
Yes there is a problem.
Explanation:
Yes there is a problem.
Looking at /22, the host part has 10 bits(right most 10 bits). The subnet mask is:
1111111.11111111.11111100.00000000
The first IP address which is 192.168.68.0 in binary form:
11000000.10101000.1000100.00000000
Here all the 10 bits of host part are 0's. Hence it is a valid network IP address.
192.168,68.0 - This is avalid subnet id
The Next IP address 192.168.69.0 in binary form:
11000000.10101000.1000101.00000000
Here all the 10 bits of host part are not 0's. Hence it is not a valid network IP address for /22 cidr.
Moreover this address lies in the first subnet - 192.168.68.0, which has host addresses in the range of 192.168.68.1 to 192.168.71.254
Hence other IP addresses 192.168.69.9/22, 192.168.70.0/22 and 192.168.71.0/22 lies n the subnet 192.168.68.0/22.