Processor Speed = 600MHZ = $6*10^{8}$ cycles/second
Transfer time = $\frac{20KB}{10MB}$ = 2ms = $12*10^{5}$ cycles
Total cycles consumed = 1200,000 + 300 + 900 = 120,1200 cycles
% of processor time consumed = $\frac{120,1200*100}{6*10^{8}} = \frac{12012}{60000}$= 0.2%
So, (D) can be chosen as option!