Please see if this approach is right or wrong.
Let Logical Address Space be x words.
Now, Page Size = Frame Size = $\frac{Physical Address Space}{No. of Frames}$ = $\frac{2^{27}}{2^{14}}$ = 213 words.
Number of Pages = $\frac{x}{2^{13}}$
Page Table Entry = Number of bits to address each frame = 14bits = 1.75B
Number of pages in inner Page Table = $\frac{x * 1.75}{2^{13} * 2^{12}}$ (Each page of a Page Table is 4KW)
Now, $\frac{x * 1.75}{2^{13} * 2^{12}}$ = 212 (Page Table is divided into 212 Pages)
Solving, we get, x = 1.14 * 236 $\approx$ 21 * 36 = 237 = 37bits.
Please check this!!