Answer:
We have:
$15000\;\text{rpm} = 60 \;\text{seconds}$
$\therefore \;1\;\text{rpm} = \frac{60}{15000} = 0.004 \;\text{seconds} = 4\;\text{msec}$
We know that:
Disk Latency = average seek time + average rotational delay + transfer time + controlled overhead.
Now, average access time neede to read $\mathbf k$-bytes $=8 + \frac{4}{2} + \frac{\mathbf k}{262144}\times 4$
For $1-\text{KB}$ block access time $=10.015625\;\text{msec}$
For $2-\text{KB}$ block access time $=10.03125\;\text{msec}$
For $3-\text{KB}$ block access time $=10.0625\;\text{msec}$
So, there is not much difference between the access time of all three.
Answer: $\therefore $ Data rates are: $102,240\;\frac{\text{KB}}{\text {sec}}, \; 204, 162\;\frac{\text{KB}}{\text {sec}}, \;\text{and} \;407,056\;\frac{\text{KB}}{\text {sec}}$