Prerequisite: The visualisation of multidimensional arrays.
How do we get to $A[i][j]$ ? (in RMO)
From the base address, we skip $i$ rows and $j$ columns. This is quite well known.
See this or this for example.
Actually, a better perspective is to see it as skipping $i$ arrays and $j$ elements.
An even better perspective is to look at it as skipping $i$ $1D$ arrays, and $j$ elements.
By extending this, $A[i][j][k]$ is nothing but skipping $i$ $2D$ arrays, $j$ $1D$ arrays and $k$ elements. You can extend this generalised perspective to any number of dimensions.
Now,
t0 = i ∗ 1024
t1 = j ∗ 32
t2 = k ∗ 4
This is nothing but "skipping". Start from the innermost index. We're skipping $k$ by multiples of $4$. So, single-elements in the array occupy space of $4$ (Bytes). Hence, we're dealing with ints.
We're skipping $j$ by multiples of $32$. This means the $1D$ "sub-arrays" are of size $32$ (Bytes).
//Visualisation of multidimensional arrays is needed. If you don't know how to picturise it, I'll draw that in the comments.
Size of the $1D$ array = $32$ .
Number of elements = $\frac{32}{4}=8$
So, the last array subscript has a size 8.
With this knowledge, Option A is the answer.
But let's continue.
We're skipping $i$ by multiples of $1024$. This means the $2D$ "sub-arrays" are of size $1024$ (Bytes).
Number of elements in $2D$ subarrays = $\frac{1024}{32}=32$
Hence, there are $32$ $1D$ subarrays.
So the middle subscript is of size 32.
Nothing can be concluded about the first subscript. (Why? Hint: we don't know the total number of 2D subarrays)
So, we conclude the array must have been declared like: $int X[?][32][8]$