total instruction size= 16 bits
so the total no. of encodings possible=2^16=65536
no of encodings with 2 address instruction=(2)*(2^7)*(2^7) =2^15=32768 (we have used 2^7 twice as it is 2-address instruction, and 7 is the size of the address field given in the question )
no. of encodings with 1-address instruction= 250*2^7=32000
now, no of 0-adress instruction=65536-(32768+32000) =768
so, the answer is (c) 768