Answer
$\approx 12\;\rm cm$
Work Step by Step
We need to put the small ball at some place on the same line ($x$-axis) that makes the net force exerted on it by the two other balls is zero.
This means that we need to put ball C between the two balls (A and B) since each ball will attract ball C in opposite direction.
Ball A attract C to the left while ball B attract C to the right.
If we put C left to ball A the net force exerted on it will be toward the right and if we put it right to ball B, the net force exerted on it will be toward the left. Hence, we should put it between the two balls A and B, as we see in the figure below.
$$\sum F_{x,\rm on\;C}=F_{g,\rm B\;on\;C}-F_{g,\rm A\;on\;C}=m_Ca_x=m_C(0)=0$$
Hence,
$$F_{g,\rm B\;on\;C}=F_{g,\rm A\;on\;C}$$
$$\dfrac{ \color{red}{\bf\not}Gm_B \color{red}{\bf\not}m_C}{r_{BC}^2}=\dfrac{ \color{red}{\bf\not}Gm_A \color{red}{\bf\not}m_C}{r_{AC}^2}$$
$$\dfrac{m_B}{r_{BC}^2}=\dfrac{m_A}{r_{AC}^2}$$
From the figure below, we can see that $r_{AC}=x$ and $r_{BC}=0.20-x$
$$\dfrac{m_B}{(0.20-x)^2}=\dfrac{m_A}{x^2}$$
Plugging the known;
$$\dfrac{10}{(0.20-x)^2}=\dfrac{20}{x^2}$$
Hence,
$$x^2=2 (0.20-x)^2=2 (0.04-0.4x+x^2)$$
$$x^2 =0.08-0.8x+2x^2 $$
$$x^2-0.8x+0.08=0$$
Thus, whether $x=0.682843$ m which is not reasonable since we need $x$ to be 0 m to 0.20 m; OR,
$$x=0.117157\;\rm m=\color{red}{\bf 11.7}\;cm$$
