A computer system supports a logical address space of $2^{32}$ bytes. It uses two-level hierarchical paging with a page size of 4096 bytes. A logical address is divided into a $b$-bit index to the outer page table, an offset within the page of the inner page table, and an offset within the desired page. Each entry of the inner page table uses eight bytes. All the pages in the system have the same size. The value of $b$ is ____________ . (Answer in integer)

1. Logical Address Space Size:

   $$\text{Logical Address Space} = 2^{32} \text{ bytes} \implies \text{Logical Address} = 32 \text{ bits}$$

2. Page Offset (Offset within the desired page):

   $$\text{Page Size} = 4096 \text{ bytes} = 2^{12} \text{ bytes} \implies \text{Page Offset} = 12 \text{ bits}$$

3. Offset within the Page of the Inner Page Table: Since each page table must fit into a single page frame of size $4096\text{ bytes}$:

   $$\text{Number of entries per inner page table} = \frac{\text{Page Size}}{\text{Inner Page Table Entry Size}} = \frac{4096 \text{ bytes}}{8 \text{ bytes}} = 512 = 2^9$$

   Therefore, the number of bits required to index within an inner page table page is $9\text{ bits}$.

4. Outer Page Table Index ($b$):

   $$\text{Total Logical Address Bits} = b + (\text{inner page index bits}) + (\text{page offset bits})$$

   $$32 = b + 9 + 12$$

   $$b = 32 - 21 = 11$$

Thus, the value of $b$ is 11.