Given : $32$ bits as disk block address size $\therefore$ conclude that there are $2^{32}$ blocks in the disk.
Now for a free space bit vector considering maximum possible file size.
Total space for free blocks = $2^{32}$ bits $= \ 2^{29}$ bytes
Total blocks we need to store this info $=\Large {\frac{2^{29}}{2^{12}}}$ $=2^{17}$ blocks
Answer given as : $2^{17}= \color{Red} {1024 \times 128}$
