Given that,
Page Size = 8KB = 213B.
Physical Address Space = 32bits.
Number of frames in main memory = $\frac{2^{32}}{2^{13}}$ = 219.
So, number of bits to address each frame = 19.
Page table entry contains 1 valid bit and 19 bits for address translation, ie 20 bits or 2.5B.
Now given, Page Table Size = 20MB = 20 * 220B.
Page Table Size = Page table entry size * Number of pages
So, Number of pages = Page table size/Page table entry size = $\frac{20 * 2^{20}B}{2.5B}$ = 223.
Logical Address Space = Number of Pages * Page Size = 223 * 213 = 236B
So, 36 bits are required to address a word of Logical Address Space.