In a disk all tracks have equal capacity and so data density is highest for the innermost track as it has the smallest radius.
- Maximum storage density (hence of innermost track) $=1400$ bits per cm
- Track capacity $\pi \times d \times 1400$ bits $ = 3.14 \times 10 \times 1400 = 43960$ bits
With $4200$ rotations per minute, data transfer rate $=\dfrac{4200 \times 43960}{60}$ bits per second.
Therefore, to transfer $64$ bits time required $ = \dfrac{60}{4200 \times 43960}\times 64 = 20.798\mu s$
With $1\mu s$ memory cycle time, the disk will take one memory cycle out of $21+1 = \dfrac{1}{21+1} \times 100 \approx 5\%$
(PS: If we consider just one word transfer we add the memory cycle time to the disk transfer time in the denominator but for continuous DMA transfer, this is not required as when data is transferred to main memory, disk can continue reading new data)