The transferable unit in disks are blocks .
Given that a complete harddisk is of size $40$ MB which is basically collection of many blocks.
So,
Step $1)$ No. of blocks which is $\Large \frac{Total \ size}{size \ of \ a \ block}$ $=\Large \frac{40MB}{8KB}$ $=5$K blocks
Step $2)$ In free space mgmt every block is assigned a bit so total no of bits needed $=5$ Kb$\ =\color{Red}{5 \times 1024}$