Suppose the object is placed at point P on the line EM joining the centres E, M of Earth and Moon. Let distance $PM=r$. Then distance $EP=R-r$ where $R=384,400km$ is the distance $EM$.

Suppose the mass of the Earth is $m$. Then the mass of the Moon is $\mu m$ where $\mu=0.0123$. The gravitational forces on an object of mass $m_0$ at point P from the Earth and Moon are respectively $$F_E=G\frac{mm_0}{(R-r)^2}$$ $$F_M=G\frac{\mu m m_0}{r^2}$$

We need to find a distance $r$ such that $$F_M=\frac13 F_E$$ $$G\frac{mm_0}{(R-r)^2}=\frac13 G\frac{\mu m m_0}{r^2}$$ Rearranging : $$\frac{3}{\mu}r^2= (R-r)^2$$ $$kr=R-r$$ $$r=\frac{R}{k+1}$$

Here $$k=\sqrt{\frac{3}{\mu}}=\sqrt{\frac{3}{0.0123}} \approx 15.617$$ hence $$r=\frac{384,400}{16.617} km\approx 23,100 km$$

**Comments :**

The fewest number of significant figures in the data is 3 so the answer is not reliable to more than 3 significant figures.

This calculation assumes that P lies outside of the Moon and the Earth. Inside the Moon and Earth the $1/r^2$ law does not apply; instead the gravitational force is proportional to $r$. Another method would have to be used. So check that the calculated value of $r$ is greater than the radius of the Moon. It is. Phew.