Solution
The correct answer is option 4
Explanation:Data Given:Tag field =9 bits
index field=12 bits
Main Memory address
total bits= 9+ 12 =21 bits
Calculation:
Size of Main memory
= 221=21 * 220
=2*1024K
=2048k
Size of cache memory
= 2index+offset
=212+0
=212
=4k