total cycle which dma need to initiate and then complete the total cycles that will be required will be 1200 cycles. Total time by dma will be ( initiate + transfer + complete ) out of which only transfer time will be useful time.
so wastage time = 1200 cycles * time of one cycle.
1cycle = $\frac{1}{frequency}$
1 cycle= $\frac{1}{600Mhz}$
1200 cycles = 2 micro second time.
now we have to calculate the time for transfer which can be easily calculated using the disk parameters given . assuming a movable read write head is available . if it is movable in one round i will be able to read one track only else if it is fixed . every surface will have a read write head and 16 tracks can be read in one rotation.
3000 rounds = 60 seconds
1 round = $\frac{1}{50}$
total data on track = 1024 *1KB
1024*1KB data = $\frac{1}{50}$ seconds
20 KB data = $\frac{20}{50*1024}$= 390 micro second.
so total time = 392 micro second. whle transfer time is 390.
cpu will be involved in the transfer time only . so percentage will be $\frac{2*100}{392}$