How many bits are needed to represent an instruction in a computer that has 53 different instructions, all of which are two addresses, when the computer has 65000-word memory ?
Don't know if I got you right, "all of which are two addresses"
Let's assume the wanted instruction may look like this <Instruction ><Operator1><Operator2>
you say there's 53 instructions, so let's create a list, and check how many bits are used:
number instruction
00000000 operation1
00000001 operation2
00000010 operation3
...
00110100 operation52
00110101 operation53
as you see, only 6 bits of the eight are used, which makes sense, 2**6 = 64
to address 65000 words, you need 16 bit (2**16=65536)
to encode an instruction with 2 operants, 16 bit each, you'd need
6 (for the instruction) +16 (op1) +16 (op2) = 38 bit