Size of Main Memory is 64 words i.e 2^6 ,So out of 16 bits 6 bits for Address and 10 bits for Opcode in 1-add format , total one add
instruction = 2^10= 1024.
Since there are two 2 add instructions , so again from 10 bits of opcode , 6 bits will be taken out, so remaining 4 bits will be left for opcode.
Now , to accomodate two 2 add instructions 2^6 one add instructions are to be satisfide,
so ,valid 1 add inst + non valid 1 add inst= total 1 add inst
valid1 add inst= 2^10 - 2* 2^6 = 1024-128= 896.