I hope you are trying to find the number of bits for all the mentioned terms. Assuming word size to be $1$ $Byte$, as nothing is mentioned about it.
Page size is $8KB$ = $2^{3+10}$ = $2^{13}Bytes$. So in page offset $13$ $bits$ are there.
Virtual memory(or logical memory) is $16 GB$ = $2^{4+30}$ = $2^{34}Bytes$. So total bits in virtual address is $34 bits$.
This means that page number has $34 -13 = 21 bits $.
Physical memory is $4GB$ = $2^{2+30}$ = $2^{32}Bytes$. This means that physical address is $32 bits$.
Answer would change is word size is NOT 1 Byte.