Total Memory = 2 ^ 18 * 32 bits =>2 ^(18+5)=2^23 bits => 2^23/2^3 B ( given byte addressable) =>2^20 B
Address length for Memory =20 bits
Mode = log 7= 3 bits
Registers= log 60 =6 bits
Opcode=32-(3+6+20)=> 32-29 =>3
So no of instructions = 2^3 = 8 diff instructions are possible