Let the page size be $x$.

Since virtual address is $46$ bits, we have total number of pages $ = \frac{2^{46}}{x}$

We should have an entry for each page in last level page table which here is $T3$. So,

Number of entries in $T3$ (sum of entries across all possible $T3$ tables) $ = \frac{2^{46}}{x}$

Each entry takes $32$ bits $= 4$ bytes. So, total size of $T3$ tables $= \frac{2^{46}}{x} \times 4 = \frac{2^{48}}{x}$ bytes

Now, no. of $T3$ tables will be Total size of $T3$ tables/page table size and for each of these page tables, we must have a $T2$ entry. Taking $T3$ size as page size, no. of entries across all $T2$ tables $= \frac{\frac{2^{48}}{x}}{x} = \frac{2^{48}}{x^2} $

Now, no. of $T2$ tables (assuming $T2$ size as page size) = $\frac{2^{48}}{x^2} \times 4$ bytes = $\frac{\frac{2^{50}}{x^2}} {x} = \frac{2^{50}}{x^3}$.

Now, for each of these page table, we must have an entry in $T1$.

So, number of entries in $T1$ $=\frac{2^{50}}{x^3}$

And size of $T1$ $=\frac{2^{50}}{x^3} \times 4 =\frac{2^{52}}{x^3} $

Given in question, size of $T1$ is page size which we took as $x$. So,

$x = \frac{2^{52}}{x^3}$

$\implies x^4 =2^{52}$

$\implies x = 2^{13}$

$\implies x = 8 \text{ KB}$