Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric flux through a surface is given by
$$\Phi=\vec E\cdot \vec A=EA\cos\theta$$
The area of the base of the tetrahedron, as we see in the figure below, is given by $A_{\rm base}=0.5L h$ where $h=\sqrt{L^2-(0.5L)^2}=\sqrt{3} L/2$. Hence, $A=\frac{\sqrt3}{4}L^2$
Thus,
$$\Phi_{\rm base} =\frac{\sqrt3}{4}EL^2\cos180^\circ$$
Plug the known;
$$\Phi_{\rm base} =-\frac{\sqrt3}{4}(200)(0.2)^2 $$
$$\Phi_{\rm base} =\color{red}{\bf -3.46}\;\rm N\cdot m^2/C$$
$$\color{blue}{\bf [b]}$$
Since the tetrahedron has three equilateral triangle sides. and since the electric field is uniform and pointing upward, the three sides will get the same amount of flux.
Hence,
$$\Phi_{\rm tetrahedron }=\Phi_{\rm base}+3\Phi_{\rm sides}$$
Thus,
$$\Phi_{\rm sides}=\dfrac{\Phi_{\rm tetrahedron }-\Phi_{\rm base}}{3}=\dfrac{0-\Phi_{\rm base}}{3}$$
where $\Phi_{\rm tetrahedron }=0$ since it has no inner charge.
Plugging the known;
$$\Phi_{\rm sides}=\dfrac{ -( -3.46)}{3}=\color{red}{\bf 1.15}\;\rm N\cdot m^2/C$$
Each side will be affected by this amount of electric flux.
