1. Use Gauss's law in differential form

For electric flux density $\vec D$,

$\nabla \cdot \vec D = \rho_v$

where $\rho_v$ is the volume charge density.

Given,

$\vec D = e^{-x}\sin y\,\hat i - e^{-x}\cos y\,\hat j + 2z\,\hat k \quad \text{C/m}^2$

2. Compute the divergence

$\nabla \cdot \vec D = \frac{\partial D_x}{\partial x} + \frac{\partial D_y}{\partial y} + \frac{\partial D_z}{\partial z}$

Here,

$D_x = e^{-x}\sin y, \quad D_y = -e^{-x}\cos y, \quad D_z = 2z$

Now,

$\frac{\partial D_x}{\partial x} = \frac{\partial}{\partial x}(e^{-x}\sin y) = -e^{-x}\sin y$

$\frac{\partial D_y}{\partial y} = \frac{\partial}{\partial y}(-e^{-x}\cos y) = e^{-x}\sin y$

$\frac{\partial D_z}{\partial z} = \frac{\partial}{\partial z}(2z) = 2$

So,

$\nabla \cdot \vec D = -e^{-x}\sin y + e^{-x}\sin y + 2 = 2$

Thus,

$\rho_v = 2\ \text{C/m}^3$

3. Find charge in the given incremental volume

Given volume,

$\Delta V = 2 \times 10^{-9}\ \text{m}^3$

Hence enclosed charge,

$Q = \rho_v \Delta V = 2 \times 2 \times 10^{-9} = 4 \times 10^{-9}\ \text{C}$

4. Convert to nC

Since $1\ \text{nC} = 10^{-9}\ \text{C}$,

$Q = 4\ \text{nC}$

5. Final answer

$\boxed{4}$

The derived answer matches the stored correct answer.