$Percentage\ of\ CPU\ time\ consumed,P = \frac{No.\ of\ CPU\ cycles\ required\ by\ DMA\ per\ second}{No.\ of\ CPU\ cycles\ available\ per\ second}$
$No.\ of\ DMA \ transfer\ requests/sec = \frac{Data\ transfer\ rate\ of\ IO\ device }{Size\ of\ one\ transfer}$
$=\frac{10MB/s}{20KB} =\frac{10*10^3}{20}= 500s^{-1}$
$No.\ of\ CPU\ cycles\ required\ by\ DMA\ per\ second$
$ = (no.\ of\ DMA\ transfer\ requests/sec)\ *\ (CPU\ cycles\ required/transfer)$
$=500 * (300+900) = 500 * 1200 = 6*10^5s^{-1}$
$No.\ of\ CPU\ cycles\ available\ per\ second = 600 * 10 ^6=6*10^8s^{-1}$
$P = \frac{6*10^5}{6*10^8}*100\%=0.1\%$
Answer : D