$2000$ $KB$ is transferred in $1$ second
$4$ $KB$ transfer is $(4/2000 ) * 1000 \text{ ms} = 2 \text{ ms}$
Total cycle required for locking and handling of interrupts after DMA transfer control
$=(1000+500) \text{ clock cycle } = 1500 \text{ clock cycle }$
Now, $50$ $Mhz = 50 * 10^6 = 0.02 \text{ microsecond}$
So, $(1500 * 0.02 ) = 30 \text{ microsecond}$
$30 \mu s$ for initialization and termination and $\ 2 ms$ for data transfer.
The CPU time is consumed only for initialization and termination.
Fraction of CPU time consumed $=\dfrac{30\mu s}{(30\mu s+2\,ms)}=0.015$