How many bits for offset/frame/page
The frame size is 2KB. Assuming memory is byte-addressable, we need an offset into 2000 different bytes. 2000 is approximately (2^10)*2 = 2^11, so we need 11 bits for the frame offset.

Then, we can easily calculate that 33 - 11 = 22 bits are used to identify  a physical page (frame), and 34 - 11 = 23 bits are needed to identify a virtual page.