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 pagesize) = $\frac{2^{48}}{x^2} \times 4$ 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,

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

$x^4 =2^{52}$

$x = 2^{13}$

$ = 8KB$