Looks like there isn't enough information to answer this question.
See, the page table size is equal to the number of entries in it times the size of each entry.
First, I don't know if the page table entries are 24-bit-long or 32-bit-long or bigger. That's not explicitly specified.
Second, if there's only one page table involved in translation of virtual addresses into physical addresses, a page table has to cover the entire 32 bits of the virtual address. Since a single PTE only handles 13 (because the page size is 2^13=8192 bytes), you'd need 2^(32-13)=2^19 entries in this case.
Now, if there's a hierarchy of page tables involved in the translation, in other words, the entries of the first page table (often called page directory) point to a second-level page table and those either point to the code/data pages or to a third-level page table and so on, you get the idea... in this case the answer is going to be different. But it's not specified whether there's just one page table or a chain of multiple page tables.
In my opinion, a page table of 2^19 entries is unlikely. A page table is usually the size of a single page (sometimes smaller). So, if, for example, PTEs are 32-bit, you have 8192/4=2048 PTEs in a PT and that covers only 11 (2^11=2048) out of the remaining 32-13=19 bits of the virtual address. The other 19-11=8 bits of the virtual address must be taken care of by another page table, page directory, having only 2^8=256 entries, provided the entries are the same in the page table and page directory.
Given the provided information, there's no definitive answer here, only guesses.
