By multiplying the below matrix element wise, we can find the answer as $A.B = C$
$$ \begin{bmatrix} 2 \\ 3 \\ \end{bmatrix}. \begin{bmatrix} 2 & 1 \\ 3 & 4 \\ \end{bmatrix} = \begin{bmatrix} 7 \\ 18 \\ \end{bmatrix} $$
Is there any way to to find B (2x2 matrix) if we know C and A?
There is a way, but you won't get a unique matrix $B$. Namely, if:
$$B=\left[\begin{matrix}a&b\\c&d\end{matrix}\right]$$
then your equation for $B$ looks like this:
$$\left[\begin{matrix}a&b\\c&d\end{matrix}\right]\left[\begin{matrix}2\\3\end{matrix}\right]=\left[\begin{matrix}7\\18\end{matrix}\right]$$
which is equivalent to the following system of equations:
$$\begin{array}{rcl}2a+3b&=&7\\2c+3d&=&18\end{array}$$
which doesn't have a unique solution. For example, you can freely choose $a$ and $c$, and set $b=\frac{1}{3}(7-2a)$ and $d=\frac{1}{3}(18-2c)$, so:
$$B=\left[\begin{matrix}a&\frac{1}{3}(7-2a)\\c&\frac{1}{3}(18-2c)\end{matrix}\right]=\left[\begin{matrix}0&7/3\\0&6\end{matrix}\right]+a\left[\begin{matrix}1&-2/3\\0&0\end{matrix}\right]+c\left[\begin{matrix}0&0\\1&-2/3\end{matrix}\right]$$
for any numbers $a,c$.