Difference between word logical addressing and byte logical addressing in this exercise?
Page size => 1024 words = 2^10 words

Since it is word addressable format 10 bits will be required to identify offset within page.

There are 10 pages as 10 can't be expressed in 2's power format we need to consider 16 for identifying 10 pages i.e. 2^4=16.

So 4 bits are  required for page number.