DMA stealing mode, where does extra 40ns come from?

Question

Consider a disk with 4000 RPM rotational speed. The disk has 1K sectors on each track with 1k capacity of each sector. The disk is operating on the cycle stealing mode of DMA. It takes 50 nsec to transfer the 16 B data from disk to memory. What is the percentage of time CPU is blocked due to DMA?

This is the question 

here is the solution

 

Can you tell where the 40ns which I have circled comes from??

Comments

The circled part should be 50, as while the disk is preparing the 16B of data CPU is not blocked right? Now once the Preparation(storing the data in a data register is done) of 16B of data is done in 240nsec the next 50 nsec is used to transfer the data to the Memory by the DMA. So this 50nsec is when CPU is blocked.

So percentage of time CPU is blocked while DMA transfers 16B of data

= actual time / total time = (50/290)*100 = 17.24%

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it should be 50 nsec not 40 nsec .

cpu transfer time/(cpu transfer time from buffer to mainmemory + data preparation time )