Because each thread hangs vertically, the net horizontal force on each ball is zero. The horizontal forces are (i) the external-field force and (ii) the mutual Coulomb force.

(1) On $A$ (positive): the attractive Coulomb force from $B$ points to the right (toward $B$); the external-field force must balance it and so points to the left. Since $Q_A > 0$, $\mathbf{E}_{\text{ext}}$ also points to the left (from $B$ toward $A$). Equating magnitudes:

$$|Q_A| E_{\text{ext}} = \dfrac{k |Q_A| |Q_B|}{r^2} \;\Longrightarrow\; E_{\text{ext}} = \dfrac{k |Q_B|}{r^2}.$$

With $r = 3\ \mathrm{cm} = 3 \times 10^{-2}$ m:

$$E_{\text{ext}} = \dfrac{(9 \times 10^{9})(2 \times 10^{-8})}{(3 \times 10^{-2})^2} = 2 \times 10^{5}\ \mathrm{V/m}, \quad \text{directed from } B \text{ to } A.$$

(2) At the midpoint $M$ (distance $r/2 = 1.5\ \mathrm{cm}$ from each ball), the fields from $Q_A$ and $Q_B$ both point in the same horizontal direction --- from $A$ to $B$ (away from the positive charge, toward the negative). Each has magnitude

$$E_{A,M} = E_{B,M} = \dfrac{k |Q|}{(r/2)^2} = \dfrac{4 k |Q|}{r^2} = 4 \times \dfrac{(9 \times 10^{9})(2 \times 10^{-8})}{(3 \times 10^{-2})^2} = 8 \times 10^{5}\ \mathrm{V/m}.$$

So the charges contribute $E_{\text{charges}} = 1.6 \times 10^{6}$ V/m from $A$ to $B$, while $E_{\text{ext}} = 2 \times 10^{5}$ V/m points from $B$ to $A$. The total is

$$E_M = 1.6 \times 10^{6} - 2 \times 10^{5} = 1.4 \times 10^{6}\ \mathrm{V/m}, \quad \text{directed from } A \text{ to } B.$$