Since the last 8 bits of the virtual address denote the page offset, we can find that page size is $2^{8}$ bytes.
Total memory is 256KB. So number of pages is $\frac{2^{18}}{2^{8}}=2^{10}=1024$.
The first 10 bits of virtual address means, we have $2^{10}=1024$ entries in the outer page table. Similarly, next 8 bits in the virtual address help us to find that there are $2^{8}=256$ entries in the second level page table, and next 6 bits in the virtual address shows us that there are $2^{6}=64$ entries in the third level page table.
Size of a page table entry is 2 bytes for all levels.
The outer page table points to 1024 entries in the second level page table each of whcih points to 256 entries in the third level page table each containing 64 pages. The process's address space consists of 1024 pages. So we need 16 third level page tables (Here, the third level page table has only 64 entries and we need to point to 1024 pages. That's why!). Thus the page table size can be calculated as:
$(1024*2)+(256*2)+(16*64*2)=4608$ bytes.
