According to the question, the arrangement of the masses is as shown below,



Gravitational force between two masses is given as
$F=\frac{G m_{1} m_{2}}{r^{2}}$
where, $G=$ gravitational constant and $r=$ distance between them.
From the given values, we can say that force between masses at $A$ and $G$. $=$ Force between masses at $B$ and $G$. $\begin{array}{ll}\Rightarrow & F_{A}=\frac{G m_{A} m_{G}}{A G^{2}} \text { and } F_{B}=\frac{G m_{B} m_{G}}{B G^{2}} \\ \text { or } & F_{A}=F_{B}=\frac{G \times 8 \times 2}{1^{2}}=16 G...(i)\end{array}$
$\left[\because\right.$ Given, $m_{1}=m_{A}=m_{B}=8 \mathrm{~kg}, m_{2}=m_{G}=2 \mathrm{~kg}$,
$A G=B G=r=1 \mathrm{~m} \text { ] }$
From the figure, resultant of $F_{A}$ and $F_{B}$ is given as
$\begin{aligned} F_{A B} &=\sqrt{F_{A}^{2}+F_{B}^{2}+2 F_{A} F_{B} \cos \theta} \\ &=\sqrt{2 F_{A}^{2}+2 F_{A}^{2} \cos 120^{\circ}} \\ &=\sqrt{F_{A}^{2}}=\sqrt{(16 G)^{2}}=16 G \\ \Rightarrow \quad F_{A B} &=F_{A}=F_{B}=16 G \quad \text { [from Eq. (i)] } \end{aligned}$
For resultant force on $2 \mathrm{~kg}$ body to be zero, $4 \mathrm{~kg}$ body should be placed at a certain distance from $G$ such that
$\mathbf{F}_{C G}=-\mathbf{F}_{A E}$
$\Rightarrow \quad \quad\left|\mathbf{F}_{C G}\right|=\left|\mathbf{F}_{A B}\right|$
or $\frac{G \times m_{C} m_{G}}{x^{2}}=G(\mathbf{1} 6)$
where, $x=$ distance between $4 \mathrm{~kg}$ and $2 \mathrm{~kg}$ body.
Here, $m_{C}=4 \mathrm{~kg}$
$\begin{aligned}
\Rightarrow & & \frac{G \times 4 \times 2}{x^{2}} &=G(16) \\
\Rightarrow & x^{2} &=\frac{1}{2} \\
\Rightarrow & x &=\frac{1}{\sqrt{2}} \mathrm{~m}
\end{aligned}$
This means, $4 \mathrm{~kg}$ body should be placed at point $P$ on line $C G$ such that $P G=\frac{1}{\sqrt{2}} \mathrm{~m}$.