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if we consider the last level consist of only 1 page table then 3.
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Answer should be 3.

Here 44 bit virtual memory and from this 13 bit required for paging ( As 16KB/2B=16K=$2^13$ )

so  1st level page size is  $2^44/2^13$ = $2^31$

then 2nd level page size is  $2^31/2^13$ =$2^18$

then 3rd level page size is $2^18/2^13$=  $2^5$
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