Step 1: Calculate the divergence of \( \vec{E} \).
The divergence of a vector field \( \vec{E} = E_x \hat{i} + E_y \hat{j} + E_z \hat{k} \) is given by:
\[
\vec{\nabla} \cdot \vec{E} = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z}.
\]
For the given \( \vec{E}(x, y, z) = 2x^2 \hat{i} + 5y \hat{j} + 3z \hat{k} \), we calculate each partial derivative:
\[
\frac{\partial E_x}{\partial x} = \frac{\partial (2x^2)}{\partial x} = 4x,
\]
\[
\frac{\partial E_y}{\partial y} = \frac{\partial (5y)}{\partial y} = 5,
\]
\[
\frac{\partial E_z}{\partial z} = \frac{\partial (3z)}{\partial z} = 3.
\]
Thus, the divergence of \( \vec{E} \) is:
\[
\vec{\nabla} \cdot \vec{E} = 4x + 5 + 3 = 4x + 8.
\]
Step 2: Set up the volume integral.
We now need to compute the following integral over the unit cube \( V \):
\[
\iiint_V (4x + 8) \, dV.
\]
Since the limits of integration are \( 0 \leq x \leq 1, 0 \leq y \leq 1, 0 \leq z \leq 1 \), the integral becomes:
\[
\int_0^1 \int_0^1 \int_0^1 (4x + 8) \, dz \, dy \, dx.
\]
Step 3: Simplify the integral.
First, integrate with respect to \( z \):
\[
\int_0^1 (4x + 8) \, dz = (4x + 8)(1) = 4x + 8.
\]
Now, integrate with respect to \( y \):
\[
\int_0^1 (4x + 8) \, dy = (4x + 8)(1) = 4x + 8.
\]
Finally, integrate with respect to \( x \):
\[
\int_0^1 (4x + 8) \, dx = \int_0^1 4x \, dx + \int_0^1 8 \, dx.
\]
The first integral:
\[
\int_0^1 4x \, dx = 2x^2 \Big|_0^1 = 2.
\]
The second integral:
\[
\int_0^1 8 \, dx = 8x \Big|_0^1 = 8.
\]
So the total integral is:
\[
2 + 8 = 10.
\]
Step 4: Conclusion.
The value of the integral is 8. Therefore, the correct answer is (B).