Ans is c) 2 MB
20 bit address line will be $2^{20}$ unique addresses
Each unique address can store 2 Bytes of data (word size = 16 bits) since it is given that size of data bus is 16 bits(8 bits=1 Byte)
So $2^{20}$ unique addresses can store =====> $2^{20} * 2$ Bytes=====>$1024*1024*2$ B=2 MB