Each sector requires a seek,

Size of File = $8\text{KB}$

Total sectors needed for file = $\Large \frac{2^{13}}{2^9} $$ = 16$

$\text{Total time = 16 (seek time + avg rotational latency + data transfer time) }$

Time to move head from cylinder to adjacent one is 2ms, So time to move cylinder from 1 to 200 is $199 \times 2 ms.$

$\text{Avg Seek time = }$ $\Large \frac{\text{Number of Cylinders}}{3} \times 2 $$ = 133.34ms$

$\text{Avg rotational latency = } \frac{1}{2} \times \frac{60}{6000} sec = 5ms $

Hence, for 1 full rotation takes 10ms or we can say to transfer 1 track or 64 sectors it takes 10ms.

So to transfer 1 sector it will take $\frac{10}{64} ms$

$\text{Total time = 16 (133.34 + 5 + } \frac{10}{64}) = 16 \times 138.34 + \frac{10}{4} = 2215.94ms $

As Vimal pointed out that Average seek time should be $\Large \frac{\text{Max Seek time}}{3}$ i was not able to find any reliable source for it,

here OP's method was correct but incomplete and also it is almost impossible to solve for each and every track manually and then take the average of seek times. So i wrote a code for it for confirmation that this formula $\Large \frac{\text{Max Seek time}}{3}$ is valid or not.

https://ideone.com/weZQrR

Here i found out that if for moving from one cylinder to adjacent one if we need 1 time unit then for $n$ cylinders max seek time can be $n-1$ units and the formula should be $\Large \frac{\text{Max Seek time + 1}}{3}$ or $\Large \frac{\text{Number of Cylinders}}{3}$.

So i got 66.7 for 199 max seek time and multiplying by 2 because 1 time unit is 2ms for question above we get $133.34$.