The net gravitational pull!

Let the distance of the object of mass $2kg$ from object $A$ be $x$ metres.

Let the mass of object $A$ be $m$ kilo-grams; therefore mass of object $B$ is $4m$

=> It's distance from object $B$ is $10-x$ metres.

Let the force of gravitation acted on the object with mass $2kg$ by object $A$ be $F_A$ and by object $B$ be $F_B$

From newton's Universal law of gravitation we get $F_A=\frac{2Gm}{x^2}$ $F_B=\frac{2G(4m)}{(10-x)^2}=\frac{8Gm}{(10-x)^2}$

As $F_A$ and $F_B$ must be equal in magnitude $\frac{2\cancel{Gm}}{x^2}=\frac{8\cancel{Gm}}{(10-x)^2}$ $\frac{1}{x^2}=\frac{4}{(10-x)^2}$ $4x^2=(10-x)^2$ $2x=10-x$ $3x=10$ $x=\frac{10}{3}=\boxed{3.33}33...$

Let the distance of the test mass $m$ from the lighter body $A$ be $r$ m. Let the mass of $A$ be $M$ and that of $B$ be $4M$ .Then

$\dfrac{GMm}{r^2}=\dfrac{4GMm}{(10-r)^2}\implies 3r^2+20r-100=0$

$\implies r=\dfrac{10}{3}\approx \boxed {3.333}$ m. Here $G$ is the Universal Gravitational Constant.

Date of 2 Kg is not necesari .

Unic place to get g=0 is over the strait line that goes from A to B

Relation of Masses is 4 => relation of d^2 has to be too = 4 => relations of d = 2

d + 2d = 10 => d = 10/3 = 3.33....