$\begin{bmatrix} &A_{1,1} &A_{1,2} &A_{1,3} &. &. &. &. &A_{1,10} \\ &A_{2,1} &A_{2,2} &A_{3,3} &. &. &. &. &A_{2,10} \\ &.. & & & & & & & & \\ &.. & & & & & & & & \\ &A_{5,1} &A_{5,2} &A_{5,3} &. &. &. &. &A_{5,10} \end{bmatrix}$
In row major order, elements are entered row by row in an array. ie; If the base address is $0$, then $A_{1,1}$, $A_{1,2}$,$A_{1,3}$... $A_{1,10}$ are entered first to locations 0 to 9. Now $A_{2,1}$ will occupy the position 10. So in this case the element $A_{4,9}$ will occupy the location $38$.
ie; address of $i,j$ th element = $n*(i-1) + (j-1)$, where $n$ is the total number of columns.
So in general, for row major order, address of $A_{i,j}$ = Base address $+$ Size of element $*(n(i-1)+(j-1))$
Given that base address $= 100$ and size of an element $=4\space bytes$
Address of the element $A_{4,9}$
$=100+4*(10(4-1)+(9-1))$
$= 100+4*38 = 252$.
