How to calculate virtual address space from page size, virtual address size and page table entry size?
Since Logical address size is 47 bit, that means logical address space is 2^47 bytes ( assuming system is byte addressable ).

Otherwise in general if the logical address is not given then also it can be found.

Page size is 16 KB
Logical address size is 47 bit
3 levels of page tables; all have the same size
Page table entry size is 8 byte


From the information assuming the entire page is being used, 

number of entries in each page = (page size) / (page table entry size ) = (2 ^ 11)


Now consider the outermost page table. It has 2 ^ 11 entries, which means the second level page table has 2 ^ 11 pages and each page has 2 ^ 11 entries.

Going by this logic it can be derived that the number of pages in process ( logical address space ) = 2 ^ 33

And given page size = 16 KB
Therefore, the logical address space size = ( 2 ^ 33) * 16 KB = 2^47 bytes