For (A) part :
No of track $=$ Recording width/ inner space between track
Recording width $=$ (Outer Diameter $-$ Inner Diameter )$/ 2= (12-4)/2= 4 \ cm$
Therefore no. of track $= 4 \ \text{cm} / 0.1 \ \text{mm} = 400$ track
Since they have ask capacity of unformatted disk , so no $96$ bytes in $4000$ bytes would be wasted for non data purpose
Whole $4000$ is used
So, total capacity $= 400 \times 8 \times 20 \times 4000 = 256 \times 10^6 \ \text{Bytes} = \mathbf{256 \ MB}$
For (B) part :
Its is given $360$ rotations in $60$ seconds
That is $360$ rotations $= 60$ sec
Therefore, $1$ rotations will take $(1/6)$ sec
In $(1/6)$ sec - we can read one track $= 20 \times (4000-96) \ B = 20 \times 3904 \ B$
Then, in $1$ sec it will be $= 20 \times 3904 \times 6$ bytes $=$ Data transfer rate $= \mathbf{ 468.480 \ KBps}$ ( when we consider $1$ Read/Write Head for all surface) .
If we consider 1 Read/Write Heads per surface ( which is default approach ), then number of surfaces = 8
Data transfer rate $= ( 468.480 \times 8 ) \ KBps = 3747.84 \ KBps $
But for our convenience we consider only $1$ surface, it reads from one surface at a time. As data transfer rate is measured wrt a single surface only .
Hence, for part B, the correct answer is $\mathbf{468.480 \ KBps}$.