Answer:
0×3604
Explanation:
Since the first instruction is byte addressable so, it will take 1 hexadecimal address so, 0×3600 is the address of printString.
Second instruction is word addressable so, it will take 2 instructions. So, 0×3601 and 0×3602 will be the address of someBytes.
Third instruction is also in byte so, it will also take 1 instruction. So, 0×3603 will be the address of moreBytes
At the end, when we will come to questionAddr uptil now 4 hexadecimal addresses will already be taken that are from 0×3600 to 0×3603 so, questionAddr will take address 0×3604.