The following data segment starts at memory address 0x3600 (hexadecimal)
.data
printstring byte "do not add decimal to hex",0
somebytes word 19 dup(0)
morebytes byte 10, 20, 30, 40, 50, 60, 70, 80, 90
questionaddr dword ?
ignoreme word ?

what is the hexadecimal address of questionaddr?

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.

the answer is you.