Divide a network 10.10.0.0 with subnet mask 255.255.0.0 into 62 subnets. 10.10.0.0
0000 1010. 0000 1010 . 0000 0000 . 0000 0000

255.255.0.0
1111 1111. 1111 1111 . 0000 0000 . 0000 0000

Required:
a. How many bits do we need to borrow from the host portion?
b. List the lowest and the highest IP address of the first usable subnet.
c. How many usable hosts are there in total before and after subnetting?

A. 6 bits

B. lowest IP address is  10.10.0.1  and the highest IP address is  10.10.3.254

C. total number of usable hosts before subnetting = 65534

total number of usable hosts after subnetting = 65408

Explanation:

The IP address 10.10.0.0 with the subnet mask 255.255.0.0 is a class B address with 16 bits of the network address portion and 16 bits of the host portion.

To subnet the address to use 62 subnets (and to reduce the host IP address wasted) 6 bits are borrowed from the host portion of the address to give 2^6 = 64 subnets with two usable host addresses.

The first network address is 10.10.0.0 and the first usable host address of the subnet is 10.10.0.1 making it the lowest address, while the broadcast address is 10.10.3.255 and the address before it becomes the higher address in the network which is 10.10.3.254

The number of usable host of the network before subnetting is 2^16 -2 = 65534, while after subnetting is (2^(16-6) - 2) x 64 = 65408.