A)

$\text{Size of each segment} = \frac{2^{16}}{8}=2^{13}$

Let the size of page be $2^k$ bytes

We need a page table entry for each page. For a segment of size $2^{13}$, number of pages required will be

$2^{13-k}$ and so we need $2^{13-k}$ page table entries. Now, the size of these many entries must be less than or equal to the page size, for the page table of a segment to be requiring at most one page. So,

$2^{13-k} \times 2 = 2^k$ (As a page table entry size is 2 bytes)

$k=7$ bits

So, $\text{ page size } = 2^7 = 128$ bytes

B)

The TLB is placed after the segment table.

Each segment will have $\frac{2^{13}}{2^9}=2^4$ page table entries

So all page table entries of a segment will reside in the cache and segment number will differentiate between page table entry of each segment in the TLB cache.

$\text{Total segments}= 8 $

Therefore $3$ bits of tag is required

C)

$\text{Number of Pages for a segment} =\frac{2^{16}}{2^9} =2^7$

$\text{Bits needed for page frame identification} = 7 \text{ bits} \\ +1 \text{ valid bit} \\ +3\text{ page protection bits} \\ +1 \text{ dirty bit} \\= 12 \text{ bits needed for a page table entry}$

$\text{Size of each page table entry} = 2$ bytes $=16$ bits

$\text{Number of bits left for aging} = 16-12 = 4$ bits