How to calculate the size of a page in a two level paging CPU?
The thorough solution:


The virtual address looks like this:  

+-------------+------------+--------+
+ table index + page index + offset +
+-------------+------------+--------+


where:


table index is the index length of an entry in the page table directory. I'll call it $i_t$.
page index is the index length of an entry in the page table. I'll call it $i_p$.

So we have such an equation now:
$$
i_t+i_p+o=32
$$
where $o$ represents bits that we need for offset.

From the decoded addresses we know that offset cannot be greater than $14$, so:
$$
o \leq 14
$$  
We know that an entry in both page tables and page tables directories needs $i_t + b_d$ of bits where $b_d$ is the number of bits for accounting information (such as dirty bits, protection bits and so on). So, $b_d$ are accounting bits in the page table directory and $b_t$ are are accounting bits in page table.  

So, if $\text{page size} = 2^\text{offset}$ then $\text{page table size} = 2^\text{offset}$ and $\text{page table directory size} = 2^\text{offset}$:
$$
\begin{cases}
i_t+b_d = o\\
i_s+b_t = o\\
\end{cases}
$$
We can say thatSince $\text{page table size} = \text{page table directory size}$ then $i:= i_t = i_s$ and $b :=b_d=b_t$, so we end up with a system of equations like this:
$$
\begin{cases}
i_t+b_d = o\\
i_s+b_t = o\\
i_t+i_p+o=32
\end{cases}
$$
$$
o-b_d+o-b_t+o=32\\
$$
and finally because of $b :=b_d=b_t$:
$$
3o-2b=32\\
$$
Now we will test different values of $o$ (I'll not test odd $o$'s because $o$ must be dividable by $2$):


when $o=14$:
$$
42-32=2b=10\\ 
\implies b=5\\ 
\implies i = 14-5=11\\
\implies i_t+i_s+o=11+11+14=36 \neq 32
$$
when $o=12$:
$$
36-32=2b=4\\ 
\implies b=2\\ 
\implies i = 12-2=10\\
\implies i_t+i_s+o=10+10+12=32 = 32
$$
which seems to be the answer. Let's just make sure there are no other possibilities.
when $o=10$:
$$
30-32=2b=-2\\ 
\implies b=-1\\ 
$$
which cannot be true - you obviously cannot have a negative number of bits.



Therefore, the size of a page is equal $2^o=2^{12}$.