Memory size = 2 i KB = 2 (i+10) byte
so, (i+10) bits are required for 1 memory cell .
here , it is given that "2j " instructions are present , so there must be "2j " opcode . so j bits for opcode .
intruction format is: !! opcode !! address !! address !!
so, total bits needed = J + (i+10) +(i+10) = J +2i + 20 bits
option c is correct.