Each datablock will have its entry.
So, Total Number of entries in the FAT $= \large \frac{\text{Disk Capacity}} {\text{Block size}} = \frac{100MB}{1KB} = 100K$
Each entry takes up $4B$ as overhead
So, space occupied by overhead = $100K \times 4B = 400KB = 0.4MB$
We have to give space to Overheads on the same file system and at the rest available space we can store data.
So, assuming that we use all available storage space to store a single file = Maximum file size = $\text{Total File System size} -\text{Overhead} = 100MB - 0.4MB = 99.6 MB$