Instruction memory has one input which is the address line $($called addr with $8$ bits$),$ and one output

which is the instruction we read from the memory $($instruction$,$ has $32$ bits$).$ With the $8$ bits address

which comes from the P C$,$ we can address $28$ bytes $($this memory is byte addressable$).$ An instruction

has $4$ bytes $(32$ bits$).$ So we can store $28/4=64$ instructions in our instruction memory. Hence, our

instruction memory is $64\backslash$times $32$ bits