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 $2^18/2^13$= $2^5$