Ans will be none
The reason behind this is that 1-address type instruction contains two-part opcode and operand.
opcode will require a decoded form of how many instructions were given ie. $\log_2i$
operand will be decoded form of size of memory ie. $log_2({2^m} KB) = \log_2(2^{m + 10}) = m + 10$
Therefore, length of instruction is $\log_2i + m + 10$ bits $=$ $\frac {(\log_2i + m + 10)}{8}$ bytes