Note that the broadcast address is the last address.
163.93.63.255 = 10100011.01011101.00111111.11111111. The default mask for class B is 255.255.0.0/16
You may note that it is not a perfect class B. Rather, it is a subnet.
Let say (assume) 163.93.00111111.11111111/18 is the subnet address and 163.93.0.0/18 is the mask. I have taken /18 as the mask because 63 (00111111) is present in the third octet. But it is an assumption we can take a bigger number also.
Let's consider another example:-
Say 163.93.00111111.11111111/23 as the subnet address and 255.255.11111000.00000000/23 (or 163.93.248.0/23) as the mask. This is a valid mask because the currently chosen subnet will range from 163.93.00111000.00000000/23 to 163.93.00111111.11111111/23. The duty of mask is to produce the network id when an AND operation is applied with the address. So, apply AND operation of the mask 163.93.248.0/23 with 163.93.63.255/23. You will get 163.93.00111000.00000000/23, which the network id or the first id of our assumed subnet.
Similarly, assume that 163.93.00111111.11111111/22 as the subnet address. The mask is 163.93.240.0/22 or 163.93.11110000.00000000/22. Because when this mask is ANDed with the given address produces 163.93.00110000.00000000/22 which is the network address or the first address of the assumed subnet.
Now consider 163.93.10000000.00000000/17 as the mask. When you do AND operation with 163.93.63.255/17 it gives 163.93.0.0/17 as the answer. That means the subnet ranges 163.93.00000000.00000000/17 to 163.93.01111111.11111111/17. But the broadcast address has to be the last address. Here the last address is 163.93.63.255/17(given in question it as the broadcast address with /17 as our assumption) but should be 163.93.127.255/17. Hence a contradiction and hence it is not the answer.
Hope it helps!