you can try this approach also...
Ok let's do it.., we have given a device with speed 1MBps and it is operating in cycle stealing mode
As we know in DMA (transfer time + preparation time) both we need to consider and here preparation time for this device is
1sec --> 1MB
16Byte --> 16 μsec
Transfer time is given as 4 μsec
% of CPU blocked due to DMA = ( Transfer time / (transfer time + preparation time) ) * 100
= (4 / (4+16)) * 100
= 20
Hence option (B) is correct... 20%