To me, this is more of an engineering studies question than a physics question, but I'll give it a shot anyway.
Since the stick is in equilibrium, the clockwise torques must equal the counter-clockwise torque.
In the diagram I made, the clockwise torques were the force at 10 cm and the mass (centred at 20 cm) and my counter-clockwise torque was the mass centred at 70 cm. These may be opposite in your diagram, you'll still get the same answer.
![](https://latex.codecogs.com/png.latex?\bg_white \\$Let $m$ be the mass of the stick in kilograms.$\\\\\frac{40}{100}m\times 9.8\times 0.2+1\times 0.3=\frac{60}{100}m\times 9.8\times 0.3\\\\0.784m+0.3=1.764m\\0.98m=0.3\\m=0.31 $ kg$)