In a two level paging environment a page table is divided into $2^{12}$ pages each of size $4 KW$. The memory is word addressable, The physical address space is $128 MW$. Memory is divided into $2^{14}$ frames. The size of inner page table is __________

My solution is, since outer page table has $2^{12}$ pages and each page of size $4KW$ i.e. $2^{12}W$ so inner page table size is $2^{12} *2^{12}W=2^{24}W$

Is this correct?

And what if it was asked **What is the size of the logical address space?**

My approach is since the outer page table contains $2^{12}$ entries so we have to do $2^{12} * number$ $of$ $pages$ in the each inner page table $* Page size$

But how to find number of pages in the inner page table ?