@mo7ammedfarooq.
Given $40-bit$ for virtual addr and $16 KB$ page-size, no of pages is
$2^{40}$ / $2^{14}$
= $2^{26}$
Now, given each PTE is 48 bits in size,
Page Table size = no. of PTEs * size of PTE
= $2^{26} * 48 bits$
= $2^{26} *$ $6bytes$ ( $\because$ 48bits = 6 bytes )
= $2^{20} * 2^{6} * 6$
= $64 * 6 MB$ ( $\because 2^{20} bytes = 1 MB $ )
= $384 MB$