Answer is already given. I am showing the formal method to solve this question.
$A=$$\begin{bmatrix} 2 &2 \\ 4&9 \end{bmatrix}$
On applying $R_{2} \leftarrow R_{2} - 2R_1$
$\begin{bmatrix} 2 &2 \\ 0 & 5 \end{bmatrix}$ $\Leftrightarrow \begin{bmatrix} 1 &0 \\ -2&1 \end{bmatrix}\begin{bmatrix} 2 &2 \\ 4&9 \end{bmatrix}$
So, here Elimination Matrix $E_{21} $ is $\begin{bmatrix} 1 &0 \\ -2&1 \end{bmatrix}$
It means we can write as :-
$E_{21}A = U$
$\Rightarrow \begin{bmatrix} 1 &0 \\ -2 &1 \end{bmatrix} \begin{bmatrix} 2 &2 \\ 4&9 \end{bmatrix} = \begin{bmatrix} 2 &2 \\ 0 &5 \end{bmatrix}$
Now, Since $E_{21}A = U.$ So, $A =E_{21}^{-1} U$
$\Rightarrow \begin{bmatrix} 2 &2 \\ 4&9 \end{bmatrix} =\begin{bmatrix} 1 &0 \\ -2 &1 \end{bmatrix}^{-1} \begin{bmatrix} 2 &2 \\ 0 &5 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} 2 &2 \\ 4&9 \end{bmatrix} =\begin{bmatrix} 1 &0 \\ 2 &1 \end{bmatrix} \begin{bmatrix} 2 &2 \\ 0 &5 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} 2 &2 \\ 4&9 \end{bmatrix} =\begin{bmatrix} 1 &0 \\ 2 &1 \end{bmatrix} \begin{bmatrix} 2 &0 \\ 0 &5 \end{bmatrix} \begin{bmatrix} 1 &1 \\ 0 &1 \end{bmatrix}$
It is $A= LDU$ decomposition. So, To make it $A=LU$ decomposition where diagonal entries in U are $1$, we have to multiply $1^{st}$ $2$ matrices i.e. $L$ and $D$. Both $L$ and $D$ are lower triangular matrices, So product of $2$ lower triangular matrices should also be lower triangular.
So,
$\Rightarrow \begin{bmatrix} 2 &2 \\ 4&9 \end{bmatrix} =\begin{bmatrix} 2 &0 \\ 4 &5 \end{bmatrix} \begin{bmatrix} 1 &1 \\ 0 &1 \end{bmatrix}$
Here, $\Rightarrow L =\begin{bmatrix} 2 &0 \\ 4 &5 \end{bmatrix} and \; U = \begin{bmatrix} 1 &1 \\ 0 &1 \end{bmatrix}$
So, $l_{22} = 5$