The amount of flux passing through any surface is proportional to the number of electric field lines passing through that surface. In new situation the field lines will interact with the different parts of the cube as shown in figure.

The flux passing through the cross section of the cube,

${\mathrm{\varphi }}_1={\mathrm{EL}}^2\cdot \cos \mathrm{\theta }$

The flux passing through the upper face of the cube,

${\mathrm{\varphi }}_2=\mathrm{E}\left(\mathrm{L}\times \frac{\mathrm{L}}{2}\right)\cos \left({90}^{\circ }-\mathrm{\theta }\right)=\mathrm{E}\frac{{\mathrm{L}}^2}{2}\sin \mathrm{\theta }$

The flux passing through the lower face of the cube,

${\mathrm{\varphi }}_3=\mathrm{E}\left(\mathrm{L}\times \frac{\mathrm{L}}{2}\right)\cos \left({90}^{\circ }-\mathrm{\theta }\right)=\mathrm{E}\frac{{\mathrm{L}}^2}{2}\sin \mathrm{\theta }$

The total flux of electric field entering the cube in this position.

$\begin{array}{c}{\mathrm{\varphi }}_{\text{total }}={\mathrm{\varphi }}_1+{\mathrm{\varphi }}_2+{\mathrm{\varphi }}_3={\mathrm{EL}}^2\cos \mathrm{\theta }+\frac{{\mathrm{EL}}^2}{2}\sin \mathrm{\theta }+\frac{{\mathrm{EL}}^2}{2}\sin \mathrm{\theta }\\ ={\mathrm{EL}}^2(\cos \mathrm{\theta }+\sin \mathrm{\theta })=5000\times 0{.1}^2\left(\cos {37}^0+\sin {37}^0\right)\\ =50\times \left(\frac{4}{5}+\frac{3}{5}\right)=70\mathrm{N}-{\mathrm{m}}^2/\mathrm{C}\end{array}$