Assume Memory is Byte Addressable
Here Cache Size =$\text{256 B}$
Cache Block Size =$\text{ 8 B}$
No of blocks in cache=$\frac{256}{8}=32$
Now, $q$ pointing to $\text{4 B}$ starting with $4104$
$q\times a[i]=q\times 4=4\times 4=16B$
Hence, $1$ cache block can contain $16$ such items
So, $1$ block is enough to hold all $16$ items in $A$
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Now, $256B$ in a cache block
But according to given condition $a[64\times i]$ only $\frac{256}{64\times4}=1$ cache line is used
So, for $16$ such variable total cache miss will be $16$
Data Cache misses $A$ over $B$ will be $\frac{16\times 16}{1\times 1}=256$