Solution:

The answer is $\mathbf{(C)}$. Let $V$ be the volume we want to compute, and let $X$ be respectively the volume of the portion of the cube cut off by the plane passing through the $6$ midpoints, and $Y$ the volume of the region cut off by the plane perpendicular to the face containing $AB$. We thus have $V + X + Y = l^{3}$, where $l = 10$ is the side of the cube. One can see that $X = \frac{l^{3}}{2}$; indeed, every plane passing through the center of a cube cuts the cube into two equal parts: to prove this one can perform a central symmetry about the center of the cube and note that the plane is mapped to itself (since the point with respect to which we are performing the symmetry belongs to the plane), the cube is likewise mapped to itself (because we are symmetrizing with respect to its center), but the two parts of the cube divided by the plane are exchanged (this can easily be seen by looking at how the vertices of the cube behave). $Y$ instead is the volume of a prism of height $l$ whose base is an isosceles right triangle of side $\frac{l}{2}$, which therefore has area $\frac{l^{2}}{8}$. We thus have $Y = \frac{l^{3}}{8}$, and consequently
$$V = l^{3} - \frac{l^{3}}{2} - \frac{l^{3}}{8} = \frac{3}{8} l^{3} = 375.$$