Given, LAS=32MB=225B, PAS=256KB=218B and Page size=4KB=212B.
No. of pages in the process=$\frac{Process Size}{Page Size}$= $\frac{2^{25}}{2^{12}}$= 213 pages.
So, no. of entries in the innermost Page Table, say PT1 =213 entries.
Size of PT1=213 x PTE size.
Since PTE size is not given, assume it to contain only frame number bits. Total no of frames in Physical memory=$\frac{2^{18}}{2^{12}}$=26 frames. So frame no bits=6 bits.
Size of PT1 =213x6 bits=$\frac{2^{13}\times 6}{2^{3}}$ B= 6x210 B > Page size. So we go for another page table, say PT2 .
No of entries in PT2 = No of pages of PT1 =$\frac{PT1 Size}{Page Size}$=$\frac{6\times 2^{10}}{2^{12}}$=$\left \lceil 1.5 \right \rceil$=2. So we need 2 pages for PT1
Size of PT2 =2x6 bits=12/8 B=1.5B<Page size. So we need 1 page for storing PT2 .
No of pages in the system=No of pages of the process + No of pages of PT1 + No of pages of PT2
=213+2+1=8192+2+1=8195.
Please correct me, if I am wrong.