Consider a computer system with 57-bit virtual addressing using multi-level tree-structured page tables with $L$ levels for virtual to physical address translation. The page size is 4 KB (1 KB = 1024 B) and a page table entry at any of the levels occupies 8 bytes. The value of $L$ is ________.

To determine the number of levels $L$ in the multi-level page table, we divide the Virtual Page Number (VPN) bits across the levels:

1. Virtual Address (VA) size: $57\text{ bits}$

2. Page size: $4\text{ KB} = 4 \times 1024\text{ B} = 2^{12}\text{ B}$

   $$\text{Page offset bits} = \log_2(2^{12}) = 12\text{ bits}$$

3. Virtual Page Number (VPN) bits:

   $$\text{VPN bits} = 57 - 12 = 45\text{ bits}$$

4. Page Table Entry (PTE) size: $8\text{ bytes} = 2^3\text{ bytes}$

5. Number of entries per page table page:

   $$\text{Entries per page} = \frac{\text{Page Size}}{\text{PTE Size}} = \frac{4\text{ KB}}{8\text{ B}} = \frac{2^{12}}{2^3} = 2^9 = 512$$

6. Bits required to index each level:

   $$\text{Bits per level} = \log_2(512) = 9\text{ bits}$$

7. Number of levels $L$:

   $$L = \frac{\text{VPN bits}}{\text{Bits per level}} = \frac{45}{9} = 5$$

Thus, the value of $L$ is 5.