Memory and bits If a memory's addressability is 64 bits. What does that tell you about the size of the memory address register (MAR) and memory data register (MDR)?
 A: The maximum size of the memory that can be used in any computer is determined by the addressing scheme. 
Machines whose instructions generate 64-bit address (memory address register (MAR)) can utilize memory that contains up to 2^64 memory locations.
Given that a memory’s addressability is 64 bits, what does that tell about the size of the MAR and MDR? 
The Memory Data Register (MDR) is the register of a computer's control unit that contains the data to be stored in the computer storage (e.g. RAM), or the data after a fetch from the computer storage. It acts like a buffer and holds anything that is copied from the memory ready for the processor to use it. So, it tells us that the MDR is 64 bits. 
However, nothing can be said about the size of MAR since it depends on the number of memory locations and does not depend on the addressability (we merely have a maximum size that it can be).
A: MAR and MDR are involved and reading memory locations and writing values (numbers, data) to memory locations, respectively. In the case the OP brought up the memory address register (MAR) needs to have 64 bits to be able to accommodate the address of any memory location within the computer's memory. 
To read memory the address of a location is written to the MAR and the value contained in that address is written to the memory's data register (MDR).
To write a value to a memory location its address is first written to MAR, and the value (which in the present example is a 64-bit number) stored in MDR. So both, MDR and MAR contain 64 bits of information.
