Consider a physical address with a page frame size of 2kb. how many bits must be used to represent the page-frame offset of the physical address?

11 bits would be enough for 0 - 2047 decimal, 2048 (2k) possible values. 2048 itself would need a 12 bit, which would allow from 0 - 4095 possible values. Useful tidbit to remember: 1024, 1k is ten bits.