The logical address looks as follows:
Page Number Offset
m-n bits n bits
$$\begin{array}{|c|c|c|} \hline \text{Page Number} & \text{Offset} \\ \hline m-n\:\text{bits} & n\:\text{bits}\\ \hline \end{array}$$
The size of the page, which is also equal to frame size is defined by hardware. If the logical address is m bit long, we have n bits for offset and higher-order m- n bits for pages.
1. We have in all 8 pages. To store info. about 8 pages we require 3 bits. This means page number needs 3 bits.
2. To address 1024 (= 2^10) words in each page, we need 10 bits, since 2^10 = 1024. Thus, the offset needs 10 bits.
The total address size = number of bits for page index + number of bits for offset = 10 + 3 = 13 bits