Consider a computer system with 40-bit virtual addressing and page size of sixteen kilobytes. If the computer system has a one-level page table per process and each page table entry requires 48 bits, then the size of the per-process page table is ________ megabytes.

1. Virtual Address Space: The virtual address is $40$ bits, so the virtual address space is $2^{40}\text{ bytes}$.

2. Page Size:

   $$\text{Page Size} = 16\text{ KB} = 16 \times 2^{10}\text{ bytes} = 2^{14}\text{ bytes}$$

   Therefore, the page offset requires $14$ bits.

3. Number of Pages (Entries in Page Table):

   $$\text{Number of Pages} = \frac{2^{40}}{2^{14}} = 2^{26}$$

4. Size of Page Table Entry (PTE):

   $$\text{PTE Size} = 48\text{ bits} = 6\text{ bytes}$$

5. Total Page Table Size:

   $$\text{Page Table Size} = 2^{26} \times 6\text{ bytes} = 6 \times 2^6 \times 2^{20}\text{ bytes} = 6 \times 64\text{ MB} = 384\text{ MB}$$

Thus, the size of the per-process page table is 384 megabytes.