Given, $m=0.1 \mathrm{~kg}$



Let $x$ be the position of mass of $0.1 \mathrm{~kg}$ from its centre and $x^{\prime}$ be the position of second mass $m_{2}$ that must be suspended to the other end to prevent the rod from toppling. So,
$$
x=50-20=30 \mathrm{~cm}=0.3 \mathrm{~m}
$$
The rod will not topple if net torque on it is zero. So, balancing the moments about its centre, we get
$m_{1} g x=m_{2} g x^{\prime}$
$\Rightarrow \quad m_{2} x=0.1 \times 0.3=0.03 \mathrm{~kg}-\mathrm{m}$
From options, it is possible only in case of $(\mathrm{c})$.
$m_{2}=0.15 \mathrm{~kg}$
$x^{\prime}=\frac{0.03}{0.15}=0.2 \mathrm{~m}=20 \mathrm{~cm}$
$\therefore$ The second mass of $0.15 \mathrm{~kg}$ should be hanged at $(50+20=) 70 \mathrm{~cm}$ mark.