Let there is n process, each process waiting time 0.5.
so CPU utilization = (1 - .5n)*100 = 99
(1/2)n = 1/100, so value of n will be 6. (27 =128 so we take 6)
there is 6 processes required for 99% utilization , each process utilize 250 KB , so total memory needed = 6*250 =1500KB.
now we have already 750KB(1000-250 as os occupied ) so required memory will be 1500-750 = 750 KB