Question

a microprocssor has a data bus with 64 lins and an address bus with 32 lines the maximum number of bits that can be stored in the memory is

Comments

Okay so in the question it is given that 

The size of the address bus =32 bit

The size of the data bus=64 bit.

Now we want to calculate the number of bits the memory can store.

Here with these given information we can’t calculate the size of the main memory because we need to how the system is addressed it is word addressable or byte addressable ? what is the word size if word addressable.?We need this information before answering the question .

Now try to dig deep to understand the problem.

It is given size of address bus is 32 bit . what does it means ?

=> Basically address bus determines  the location in the memory from where the processor will read or write data .Here address space is 32 bit means we can address  $2^{32}$ unit of information.So we can think like we have $2^{32}$ number of location each location can contain some unit of data . This unit is called word size.

It is given size of data bus is 64 bit . what does it means?

=> Data bus contain the contents that have been read from memory or are to be written into the memory location . Now here size of the data bus is 64 bit so in one memory access we can access 64 bit amount of data. Now don’t confuse this with the word size .Word size is the smallest addressable unit in the memory and depends on the system architecture . Some times they can be same but not necessarily .

so what we can understood from here 

=>address bus is 32 bit so we can have $2^{32}$ possible location each location contain word size amount of data and word size depend on the system.

=> Data bus is 64 bit means in one memory access we can fetch 64 bit of data in data bus. And is not necessarily same as Word size of the memory.

Now for the completion here lets assume the word size is

1. word size =8 bit,  then we have $2^{32}$ memory location each has 8 bit of data then  size of the main memory , $2^{32}*8=2^{35}$bit=$2^{32}$ byte.(This is also gererally termed as byte addressable if word size is 8 bit =1 byte.)
2. word size =16 bit,  then we have $2^{32}$ memory location each has 16 bit of data then  size of the main memory , $2^{32}*16=2^{36}$bit=$2^{33}$ byte.
3. word size =64 bit then we have $2^{32}$ memory location each has 64 bit of data then  size of the main memory , $2^{32}*64=2^{38}$bit=$2^{35}$ byte. 

So ,Size of the memory is not depends on data bus is depends on address bus and size of the word.

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@Arjun sir in the above comment is my understanding is correct ? if there is any wrong concept please correct me.

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@kabir thanku for make me understand this concept in such a great way  my mistake was there without confirming any concept i just rushed towards the answer and not even understand the problem

Sorry for posting wrong answer