Ok Let us first calculate the time required to access the drums for one-page access.
Time to access drum once = Time to access the sector + Time to transfer the data = Avg rotational latency + Time to transfer 1000 words
Avg rotational latency = $\frac{One.Rotation.Time}{2}$ = $\frac{1}{100}seconds$ = 10,000 micro seconds
Time to transfer one page(=1000 words) = $\frac{1}{10^{3}}$ seconds = 1000 micro seconds
Therefore, time to access the drum once = 11,000 micro seconds
Now we can easily solve this problem
Every instruction need one cycle to initially access the page = 1 micro second
1% need additional one cycle = $0.01*1microsecond=0.01 micro second$
Of those 1% only 20% causes page fault = $0.01*0.2*11000 micro seconds=22 micro seconds$
Of those one percent’s 20%, 50% need to swap with a dirty page in the memory = $0.01*0.2*0.5*11000 micro seconds=11 micro secons$
Now summing up all the cases = 1+0.01+22+11=34.01 micro second ≅34 micro second