I'm preparing for the exams and this question came up -Consider a machine with $64MB$ physical memory and a $32$-bit virtual address space. If the page size is $4KB$, what is the approximate size of the page table?  (A) 16MB  (B) 8MB  (C) 2MB  (D) 24MBThe way I've solved it -Physical Address Space = $64MB = 2^{26}B$Virtual Address = $32$-bits, $\therefore$ Virtual Address Space = $2^{32}B$Page Size = $4KB=2^{12}B$Number of pages =$\,\Large\frac{2^{32}}{2^{12}}$$=2^{20}$ pages.Number of frames =$\,\Large\frac{2^{26}}{2^{12}}$$=2^{14}$ frames.$\therefore$ Page Table Size = $2^{20}\times 14\,bits \approx 2^{20}\times 16\,bits\approx 2^{20}\times 2B= 2MB.$Some books claim the answer to be $8MB$ and I don't see why, but that confuses me.Is this the correct way to solve it? Is the answer correct? 
Can this question be solved without knowing the Page Table Entry?
operating systems;memory management;paging
Your answer and calculation is correct, if we used a one-level page table.However, a one-level page table is probably not a very likely implementation strategy, so it's more interesting to analyze the space consumption for a more plausible implementation strategy and work out what this would look like.A two-level page table is probably a more likely implementation strategy.  The first 9 bits of the virtual address will be used as an index into the first-level table; they yield the entry in the second level.  The next 11 bits of the virtual address will be used as an index into the second-level table.With this strategy, the size of the entire page table is $1+k$ pages, or $(1+k) \times 2^{12}$ bytes, where $k$ is the number of second-level table entries.  The worst case is $k=2^9$, in which case the entire page table takes up approximately $2^9 \times 2^{12} = 2^{21}$ bytes, i.e., 2 MB.  In other words, the worst case for a two-level page table is approximately the same as the size of a one-level page table.  However, the best case is much less.I don't know how the books got 8 KB, but that seems clearly wrong, absent further assumptions; that would only allow 4096 entries in the entire page table, which clearly isn't enough to maintain a virtual-to-physical mapping for all of virtual memory.