The law implies that the total electric flux through a closed surface is zero if no charge is enclosed by the surface.

The electric field is uniform and we are considering a closed cylindrical surface with its axis parallel to the uniform field $\overrightarrow{\mathrm{E}}$.

$\phi_{1}$ and $\phi_{2}$ represent the flux through the surfaces $1$ and $2$ of the cylinder and $\phi_{3}$ is the flux through the curved cylindrical part of the closed surface.

Now the normal to the surface $3$ at every point is perpendicular to $\vec{E}$ so by definition of flux, $\phi_{3}=0$.

Further, the outward normal to $2$ is along $\vec{E}$ while the outward normal to $1$ is opposite to $\vec{E}$,

Hence, $\phi_{1}=-\mathrm{ES}_{1}=-\mathrm{ES} \quad\left(\because \mathrm{S}_{1}=\mathrm{S}\right)$

$\phi_{2}=\mathrm{ES}_{2}=\mathrm{ES} \quad\left(\because \mathrm{S}_{2}=\mathrm{S}\right)$

where $\mathrm{S}$ is the area of circular cross-section.

Thus, the total flux is zero, the total charge contained in the closed surface is zero.