This is an AMAZING question! It's got beautiful small details.
Data transfer rate.
To calculate that, we need track capacity.
The maximum storage density of the disk is 1400 bits/cm.
All the tracks have equal capacity. The outermost track is the largest, and the innermost track is the smallest. Still they got the same capacity, how? Because we vary the density (ie we pack the bits tightly/loosely)
The innermost track would be the most dense, so as to get the required capacity.
So, the given density is of the innermost track.
Now, what is its circumference?
$2\pi r=\pi d$
=>$10\pi $
=> $31.4cm$
1 cm of this track = 1400 bits.
31.4 cm of this track = 43960 bits. = 5495 bytes.
So, Data transfer rate = $\frac{4200(5495)}{60}$ bytes per second.
Therefore, to transfer 64-bits = 8 bytes, time required = $20.798\mu s$
Now, percent of cycles stolen:
$\frac{1 \mu s}{20.798 \mu s}$
=> $\frac{1}{21}$
=> $5\%$