An electric field given by $\overrightarrow{\mathrm{E}}=4\widehat{\mathrm{i}}+3\left({\mathrm{y}}^2+2\right)\widehat{\mathrm{j}}$ pierces Gaussian cube of side 1 m placed at origin such that its three sides represents x, y and z axes. The net charge enclosed within the cube is

The x-component of electric field is constant and does not contribute to net flux through the closed surface. Due to y-component,

Incoming flux= -6 units ( $\therefore$ y = 0 )
Outgoing flux= 9 units ($\therefore$ y = 1)
By Gauss' law, 

$\begin{array}{l}{\mathrm{\varphi }}_{\mathrm{net}}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_{\mathrm{o}}}\\ \Rightarrow \left(9-6\right)=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_{\mathrm{o}}}\\ \mathrm{Hence},\mathrm{q}=3{\mathrm{\epsilon }}_{\mathrm{o}}\end{array}$