Given,
Virtual Address Space $=57$ bits
Page size $=4$ KB =$12$ bits
Page number= $(57-12)=45$ bits
Page table entry= $8$ bytes.
Each page can contain =$\frac{4 KB}{8B}=\frac{2^{12}}{2^{3}}=2^9$ page table entries. So, we need $9$ bits to index the page table .
So, Number of levels = $\lceil \frac{45}{9} \rceil=5$ .
So ,Value of L is $5$ .