(c)

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.