Answer:
B. 80%
Explanation:
To assertain:
1 hit takes 1 clock cycle while 1 miss takes 6 clock cycles
Assuming that:
[tex]So, p + (100 - p)*6 = 200\\ \\p+600-6p = 200\\\\-5p + 600 = 200\\\\5p = 600 -200\\\\p = \frac{400}{5} \\\\p= 80[/tex]
Hence, option B. 80% will result in an effective access time of 2 clock cycles.