Given situation is
Weight of lamina acts through its centroid $G$ to prevent tilting of lamina, let a mass $m_1$ is added at vertex $B$. From $A$, perpendicular $A E$ is dropped on $B C . A D$ is medium and $G$ is centroid of $\triangle A B C$.
Now, consider $\triangle A B C$ and $\triangle E B A$.
$\triangle A B C \sim \triangle E B A$
$\frac{A B}{E B}=\frac{B C}{A B} \Rightarrow E B=x=\frac{A B^2}{B C} \Rightarrow E B=x=\frac{9}{5}$
So, $D E=B D-E B$
$\quad=\frac{5}{2}-\frac{9}{5}=\frac{25-18}{10}=\frac{7}{10} \,cm$
Now, consider $\triangle A D E, G$ is centroid of $\triangle A B C$.
So, $\quad \frac{A G}{G D}=\frac{2}{1}$ or $A G=\frac{2}{3} A D$
Also, $G H$ is parallel to $D E$.
So, $\frac{A G}{A D}=\frac{G H}{D E}$
$\Rightarrow \quad G H =\frac{A G \times D E}{A D}=\frac{\frac{2}{3} A D \times D E}{A D}$
$=\frac{2}{3} \times \frac{7}{10}=\frac{14}{30}=\frac{7}{15} \,cm$
For $B C$ to remain horizontal, torque of $m_1 g$ about $A$ must be balanced by torque of $m g$ about $A$.
$\Rightarrow m g \times G H=m_1 g \times B E$
$\Rightarrow 540 \times \frac{7}{15}=m_1 \times \frac{9}{5}$
$\Rightarrow m_1=\frac{540 \times 7 \times 5}{15 \times 9}=140 \,g$
So, mass of $140 \,g$ must be added to vertex $B$, so that $B C$ remains horizontal.
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