Answer is (c)
Class 
Leading
bits 
Size of network
number bit field 
Size of rest
bit field 
Number
of networks 
Addresses
per network 
Total addresses
in class 
Start address 
End address 
Class A 
0 
8 
24 
128 (2^{7}) 
16,777,216 (2^{24}) 
2,147,483,648 (2^{31}) 
0.0.0.0 
127.255.255.255 
Class B 
10 
16 
16 
16,384 (2^{14}) 
65,536 (2^{16}) 
1,073,741,824 (2^{30}) 
128.0.0.0 
191.255.255.255 
Class C 
110 
24 
8 
2,097,152 (2^{21}) 
256 (2^{8}) 
536,870,912 (2^{29}) 
192.0.0.0 
223.255.255.255 
Class D (multicast) 
1110 
not defined 
not defined 
not defined 
not defined 
268,435,456 (2^{28}) 
224.0.0.0 
239.255.255.255 
Class E (reserved) 
1111 
not defined 
not defined 
not defined 
not defined 
268,435,456 (2^{28}) 
240.0.0.0 
255.255.255.255

We have $32\text{ bits}$ in the $\text{IPV4}$ network
Class A $=\text{ 8 network bits + 24 Host bits}$
Class B $=\text{ 16 network bits + 16 Host bits}$
Class C $=\text{ 24 network bits + 8 host bits}$
Now for Class C we have $3\text{ bits}$ reserved for the network id...
Hence remaining bits are $21.$ Therefore total number of networks possible are $2^{21}$.
Similarly in Class B we have $2\text{ bits}$ reserved...
Hence total number of networks in Class B are $2^{14}$.
And we have $1\text{ bit}$ reserved in Class A, therefore there are $2^7$ networks.
And a better reasoning for the bit reservation is given here. have a look.
Reference: https://en.wikipedia.org/wiki/Classful_network