$\color{blue}{\underline{\text{Most Easy Solution to this problem and applicable to all such problems:}}}$
$\color{blue}{\underline{\text{Explanation:}}}$
$\mathbf{\color{red}{Row\;Major\;Order:}}$
Let's convert the above arrays to indices which starts from $\mathbf{0}$.
$\therefore[1\dots10][1\dots15]\equiv[0\dots9][0\dots14]$
Now, we need to find the position of $\mathbf{\left (i,j\right)^{th}}$ element.
$\because$ We subtracted $\mathbf{1}$ from the above given arrays to make their starting indices $\mathbf{0}$.
$\therefore \mathbf{ \left (i, j\right) } \equiv \mathbf{\left(i-1,j-1\right)}$.
Now,
$\therefore\mathbf{Address} = \mathrm{\underset{Base\;address}{100} + \underset{No.\;of\;rows}{(i-1)}\times\underset{\text{No. of columns in each row.}}{15}+\underset{\text{Number of cols. we need to traverse after rows}}{j-1} \\= 100+15\times i-15+j-1 \\=84+15\times i+j \\= 15i+j+84 }$
$\therefore\;\mathbf{15i+j+84}$ is the required answer. Hence, $\mathbf{Option-A}$ is corect.
$\color{red}{\mathbf{Column\;Major\;Order:}}$
$\color{red}{\mathbf{\underline{Swap\;column\;to\;rows\;and\;rows\;to\;columns:}}}$
Let's convert the above arrays to indices which starts from $\mathbf{0}$.
$\therefore[1\dots10][1\dots15]\equiv[0\dots9][0\dots14]$
On Swapping column to rows and rows to column:
$[0\dots14][0\dots9]$
Now, we need to find the position of $\mathbf{\left (i,j\right)^{th}}$ element.
$\because$ We subtracted $\mathbf{1}$ from the above given arrays to make their starting indices $\mathbf{0}$.
$\therefore \mathbf{ \left (i, j\right) } \equiv \mathbf{\left(i-1,j-1\right)}$.
On swapping:
$\mathbf{(j-1,i-1)}$
Now,
$\therefore\mathbf{Address} = \mathrm{\underset{Base\;address}{100} + \underset{No.\;of\;rows}{(j-1)}\times\underset{\text{No. of columns in each row.}}{10}+\underset{\text{Number of cols. we need to traverse after rows}}{i-1} \\= 100+10\times j+i-1 \\=89+10\times j+i \\= 10j+i+89}$
$\therefore\;\mathbf{10j+i+89}$ is the required answer.
$\therefore \mathbf{D}$ is the correct option in case of $\mathbf{Column\;Major\;Order}$.