The total charge enclosed in an incremental volume of 2 × 10 - 9   m 3 located at the origin is…

Electric flux density,

$\left(\overrightarrow{D}\right)=\frac{ \text{charge} }{ \text{Area} }\times \hat{\mathrm{r}}=\frac{Q}{4\pi r^2}\hat{\mathrm{r}}={ϵ}_0\left(\frac{Q}{4\pi {ϵ}_0{r}^2}\hat{\mathrm{r}}\right)$

$\Rightarrow \overrightarrow{E}=\frac{\overrightarrow{D}}{{ϵ}_0}=\frac{e^{-x}siny\hat{\mathrm{i}}-e^{-x}cos(\hat{\mathrm{j}}+2z\widehat{k)}}{{ϵ}_0}$

Also, by Gauss's law,

$\frac{\rho }{{ϵ}_0}=\left(\frac{\partial }{\partial x}\hat{\mathrm{i}}+\frac{\partial }{\partial y}\hat{\mathrm{j}}+\frac{\partial }{\partial z}\hat{\mathrm{k}}\right)·\overrightarrow{E}=\left(\frac{\partial }{\partial x}\hat{\mathrm{i}}+\frac{\partial }{\partial y}\hat{\mathrm{j}}+\frac{\partial }{\partial z}\hat{\mathrm{k}}\right)·\frac{\overrightarrow{D}}{{ϵ}_0}$

$\Rightarrow \rho =\frac{\partial }{\partial \mathrm{x}}\left(e^{-\mathrm{x}}\sin y\right)+\frac{\partial }{\partial \mathrm{y}}\left(-e^{-x}\cos y\right)+\frac{\partial }{\partial \mathrm{z}}\left(2z\right)$

$\rho =-e^{-x}\sin y+e^{-x}\sin y+2$
At origin $\rho =-e^{-0}\sin 0+e^{-0}\sin 0+2$
$\rho =2 \mathrm{C} {\mathrm{m}}^{-3}$

Charge $=\rho \times$ volume $=2\times 2\times {10}^{-9}=4\times {10}^{-9}=$
$4 \mathrm{nC}$