Answer
$$\boxed{E_{net}=\dfrac{\sqrt{3}ke}{l^2}}$$
$$\boxed{F_{net}= \dfrac{\sqrt{3}ke^2}{l^2} }$$
Work Step by Step
The electric field exerted by the left charge at the top vertex of the triangle is toward the charge itself and is equal to the electric field exerted by the right charge at the same vertex since both are at the same distance and since both are having the same charge magnitude and sign.
Therefore, and due to symmetry, the horizontal components of the two electric fields cancel each other while the vertical components are building up together downward.
Thus, noting the figure below,
$$E_{net}=E_y=E_{y1}+E_{y2}=-\dfrac{kq}{r}\cos 30^{\circ}-\dfrac{kq}{r}\cos 30^{\circ}$$
The negative signs are due to the downward direction.
Hence,
$$E_{net}=-2 \dfrac{kq}{r^2}\cos 30^{\circ}=-2 \dfrac{k(-e)}{l^2}\cos 30^{\circ}$$
$$\boxed{E_{net}=\dfrac{\sqrt{3}ke}{l^2}}\tag{downward}$$
When we place a third identical charge at the top vertex of the triangle, the net force exerted on it must be upward (since the repulsive force exerted by the two other charges are having an angle upward) while the horizontal force components are canceling each other.
Thus,
$$F_{net}=F_{y}=2eE\cos 30^\circ$$
$$F_{net}= 2e\dfrac{ke}{l^2}\cos 30^\circ$$
$$\boxed{F_{net}= \dfrac{\sqrt{3}ke^2}{l^2} }\tag {upward}$$
