let level 1 access time t1 & level 2 t2...
then t2 = 5 t1-----> eq(1)
t1 = tavg - 10====>eq2
tavg = t1 + 10
t1 = 20 ns (given)
from eq1 we get,,,
t2 =100 ns
from eq2 we get,,,
tavg = 30ns
now for simultaneous memory organization..
tavg = H t1 + (1-H) t2
30 = H *20 + (1-H) *100
H = .875