The answer is (B).
In $\mathbb{C},$ arrays are always stored in the row-major form.
Formula to evaluate $2-D$ array's location is:
$loc(a[i][j]) = BA + [(i-lb_1)\times NC+(j-lb_2)]\times c$
where,
- $\text{BA}$ - Base Address
- $\text{NC}$ - no. of columns
- $c$ - memory size allocated to data type of array $a[lb_1 \ldots ub_1] [lb_2\ldots ub_2]$
Here, $\text{BA} = 0, \text{NC} = 100, c=1, a[0 \ldots 99][0\ldots 99]$, so $lb_1=0 , lb_2=0$
$loc(a[40][50])= 0+ [ (40-0)\times 100 + (50-0)]\times 1$
$\qquad \qquad \quad = 0+[4000+50]\times 1 = 4050$.