$j-1 + i(i-1)/2$ because if you form a lower triangular matrix it contains elements in rows $1,2,3,\ldots$
So, C is the correct answer.
PS: Though not mentioned in the question, from options it is clear that the array index starts from $1$ and not $0.$
Explanation :
In a lower triangular matrix, $i^{th}$ row contains $(i +1)$ number of non zero elements.
If we assume Array index starting from $1$ then, $i^{th}$ row contains $i$ number of non zero elements.
before $i^{th}$ row there are $i -1$ rows (row $1$ to $i-1$) , and in total these rows has $1+2+3+\ldots +(i-1)= i((i-1)/2$ elements (row $1$ has $1$ element, row $2$ has $2$ elements, row $i-1$ has $i-1$ elements etc.)
Now, at $i^{th}$ row, before $j^{th}$ element there are $(j-1)$ elements(starting from $arr[i,1]$ to $arr[i,j-1]$) .
Hence, in total before $arr[i,j]$ there are $(i(i-1)/2 + j-1)$ elements and those elements will have indexes .
S,o the index of the $(i,j)^{th}$ element of the lower triangular matrix in this new representation is $(j-1) + i(i-1)/2$ which is option C .