Answer
$4.06\;\rm g$
Work Step by Step
First of all, we need to draw the force diagram of the positive charge, as shown below.
The system is stationary, so the net force exerted on each charge is zero.
$$\sum F_x=F_{\rm attractive}+T\sin10^\circ-F_{field}=ma_x=m(0)=0$$
$$F_{\rm attractive}+T\sin10^\circ-F_{field}=0$$
$$\dfrac{kq^2}{r^2}+T\sin10^\circ-qE=0\tag 1$$
We ignored the signs of the charges and focused on the charges of the force directions.
Now we need to find $r$, from the geometry of the given figure,
$$\sin10^\circ=\dfrac{\dfrac{r}{2}}{0.50}=\dfrac{r}{1}=r$$
Plug into (1) and solving for $T$,
$$\dfrac{kq^2}{(\sin10^\circ)^2}+T\sin10^\circ=qE $$
$$ T=\dfrac{qE}{\sin10^\circ}-\dfrac{kq^2}{(\sin10^\circ)^3} \tag 2$$
$$\sum F_y=T\cos10^\circ-mg=ma_y=m(0)=0$$
Thus,
$$T\cos10^\circ=mg$$
Hence,
$$m=\dfrac{T\cos10^\circ}{g}$$
Substitute $T$ from (2),
$$m=\dfrac{ \cos10^\circ}{g}\left[ \dfrac{qE}{\sin10^\circ}-\dfrac{kq^2}{(\sin10^\circ)^3}\right]$$
Plugging the known;
$$m=\dfrac{ \cos10^\circ}{9.8}\left[ \dfrac{(100\times 10^{-9})(10^5)}{\sin10^\circ}-\dfrac{(9\times 10^9)(100\times 10^{-9})^2}{(\sin10^\circ)^3}\right]$$
$$m=\color{red}{\bf 4.06}\;\rm g$$
The two spheres are identical, so both of them must have the same mass.
