Concept :
Firstly, CPU will generate logical address of the page and using this address searching in page table present in main memory to get physical frame number of page is done and then page will be accessed from main memory which will take $T_{mm}$=200ns (as given in question.)
if page is not present in Main memory then it means page fault occurs and hence Page Fault Service will be done and page will be brought from secondary memory to Main-Memory.
Now if any Main-Memory frame is free where this page can be copied it will be directly copied, but if all frames are occupied by other pages then a page replacement algorithm will run and it will select 1 page which will be removed from MM Frame so that the page requested by CPU can come inside MM.
Now suppose a page is chosen to be replaced then there are two possibilities :
(i) the page that is going out of MM is not modified that means no write operation performed in it hence it will be directly overwritten by page that CPU requested and it will take $T_{not-modified}$ = 7ms.(as given in question.)
(ii) the page that is going out of MM is modified that means write operation performed in it hence it will be firsly moved to hard disk then our requested page will come in this frame and it will take $T_{modified}$ = 15ms.(as given in question.)
also given that,
page fault rate (P) is 5% =0.05 and hence,
(1-P)= when page fault not there =0.95
Modified pages (M) is 60% =0.6 and hence,
(1-M) = when page is not modified =0.4
Now,
Effective Memory Access Time = (1-P)*($T_{mm}$) + P[ (1-M)*($T_{not-modified}$) + M*$T_{modified}$ ]
= (0.95*200ns) + 0.05[ (0.4*7ms)+(0.6*15ms)]
=(190ns)+ 0.05(2.8ms+9ms)
=(190ns)+0.05(11.8ms)
convert all to microseconds
=(0.19microseconds)+0.05(11800microseconds)
=(0.19+590)
=590.19 micro seconds
So option b) is correct.
