Memory capacity = $2^{m}KB$ = $2^{m+10}B$
Since byte addressable memory is given $1$ word = $1$ byte
Memory capacity = $2^{m+10}$ words
No. of bits required to point 1 memory location = $m+10$ bits
Total no. of operations = $2^{n}$
No. of bits to define 1 operation = $n $bits
Given an instruction have 3 operands and 1operator.
No. of bits needed by instructions = $3*(m+10) + n$
=$ 3m + n + 30$ bits
Option d) is the correct answer