Answer
Dimensions: $1024\times1024$.
MAR size: $20$ bits
Work Step by Step
1MB = $2^{20}$ bytes
The dimensions of the memory will be $1024\times1024$.
The MAR will have 20 bits to store the $2^{20}$ addresses. The row and column decoder both will have 10 bits and both decoder will give one line as output whose intersection will point to that memory location.
