Here, The curve $y=\sqrt{x}$ and the line $2y+3=x$ will intersect each other. So
$\begin{align} y&=\sqrt{2y+3}\\\Rightarrow y^2-2y-3&=0\\\Rightarrow (y-3)(y+1)&=0\\\therefore y &= -1,~3\end{align}$
From the curve $y=\sqrt{x}$, we get $y$ is always non-negative i.e. $y\ge 0$. Therefore $y\ne -1$ so only $y=3$ is solution of the equations. Now putting $y=3$ to the equation $2y+3=x$, we obtain $x=9$.
Therefore, the curve and the line intersect at the point $(9,3)$. Here is shown the graph below.
Now, the line $2y+3=x\Rightarrow y = \frac{x-3}{2}$ cuts $x$-axis at the point $(3,0)$. $\therefore$ The required region is bounded the by points $(0,0), (3,0)\text{ and }(9,3)$.
$\begin{align} \therefore \mathrm{Area} &= \int_{0}^{9}\sqrt{x}~\mathrm{d}x-\int_{3}^{9}\frac{x-3}{2}~\mathrm{d}x\\&= \left[ \frac{2}{3} x^{\frac{3}{2}}\right]_{0}^{9}-\left[ \frac{(x-3)^2}{4}\right]_{3}^{9}\\&=\left[ \frac{2}{3}(9)^{\frac{3}{2}} -0\right]-\left[ \frac{6^2}{4}-0 \right]\\&=18-9\\&=9 \end{align}$
So the correct answer is A.