Answer
$ 1.05\times 10^{-7}\;\rm C$
Work Step by Step
First, we need to sketch this problem, as seen below.
Now Let's take the point charge 1 at which the net electric force exerted on it by the two other charges is toward the north.
We assume that north here is the positive $x$-direction, as we see in the third figure.
We have here 2 directions forces components, north-south, east-west [and those are forming the horizontal plane of the equilateral triangle], upward-downward.
From the figures below, we can see that the net horizontal force exerted on charge 1 is toward the north direction while the north-south components from the two other charges cancel each other.
The vertical forces exerted on charge 1 are the tension force which has two components one of them is toward the south (the balance of the net electric force on 1) and the other is upward (that balances the weight of the ball).
The weight of the ball 1 is downward.
Thus, from the geometry of the figures below,
$$\sum F_{x}=\left[2\dfrac{kq^2}{0.2^2}\cos30^\circ\right]-T\cos\theta=0$$
So the charge is given by
$$ q =\sqrt{\dfrac{0.2^2T\cos\theta}{2k\cos30^\circ} }\tag 1$$
And
$$\sum F_{y}=T\sin\theta-mg=0$$
Thus,
$$T=\dfrac{mg}{\sin\theta}\tag 2$$
Now we need to find $\theta$, noting that there is an imaginary right triangle of 3 heads, the center of the equilateral triangle, charge 1, and the point of the ceiling at which the wire is suspended; where the wire here is the hypotenuse of this triangle.
Thus,
$$\cos\theta=\dfrac{d}{L}=\dfrac{d}{0.80}\tag 3$$
where $d$ is the hypotenuse of the right triangle in horizontal plane (see the dark yellow triangle in the second figure below) and it is given by
$$\cos30^\circ=\dfrac{0.1}{d}$$
Hence,
$$d=\dfrac{0.1}{\cos30^\circ}$$
Plugging into (3) qnd solving for $\theta$,
$$\theta=\cos^{-1}\left[ \dfrac{0.1}{0.80\cos30^\circ}\right]=\bf 81.7^\circ$$
Plugging into (2) and;
$$T=\dfrac{mg}{\sin81.7^\circ} $$
Plugging into (1) and then plug the rest known;
$$ q =\sqrt{\dfrac{0.02^2\left(\dfrac{mg}{\sin81.7^\circ}\right)\cos81.7^\circ}{2k\cos30^\circ} } $$
$$ q =\sqrt{\dfrac{0.2^2mg}{2k\cos30^\circ\tan81.7^\circ} } $$
$$ q =\sqrt{\dfrac{0.2^2 (0.003)(9.8)}{2(9\times 10^9)\cos30^\circ\tan81.7^\circ} } $$
$$q=\color{red}{\bf 1.05\times 10^{-7}}\;\rm C$$
