Electric field in a space varies with $x$as $\overrightarrow{E}=(x\hat{i}+10\hat{j})$ where $x$ is in meter. Then the total charge contained within a prism as shown is $x{\epsilon }_0$. The value of x is 

Flux through area OPQ and area RST  is zero as field perpendicular to these areas is zero.

Flux entering through area OQRS is exiting through area PQRT.

Flux entering through area POST is also zero as the field perpendicular to the area POST is zero i.e, $E_x=x \Rightarrow E_{x=0}=0.$

So, the net flux will exiting through area PQRT.

At X=x consider a dx element 

$d\varphi = E_x.\left(Area perpendicular to E_x\right)=\left(x\right).\left[2.\left(\frac{2}{3}dx\right)\right]=\frac{4x}{3}dx$

Therefore, $\varphi ={\int }_0^3\frac{4x}{3}dx={\overline{)\frac{2x^2}{3}}}_0^3=6$

$\because \varphi =\frac{q}{{\epsilon }_0}=6$

$\therefore q=6{\epsilon }_0$