for L1:-
Array size=60 x 60=3600 bytes
Block size=16 bytes
Element size=4 bytes
Number of elements in one block=4 bytes
When first element A[0][0] is referred, we bring block number 0 of main memory to line number 0.We not only bring A[0][0] element but also 22 - 1 elements(as we bring the whole block).So, when first element is referred, gives a miss. Next 22-1 elements gives hit.
Miss rate= 1/4
No. of misses= 1/4 x number of references.
No. of misses= 1/4 x 60 x 60 =900.
for L2:-
No. of Blocks in cache = 512.
The elements are already present in the cache, so no miss operation will takes place for this 512 blocks.
Number of misses=900-512=388
Total Number of misses in cache=900+388=1288