Answer: D

Small Hint:

Model the large cube as $[0,3]^3$ and write the plane through its center perpendicular to the diagonal

Big Hint:

Index each unit cube by its minimum value of $x+y+z$ and determine which index sums allow an intersection

Solution:

Take the large cube as $[0,3]^3.$ The plane is $x+y+z=\frac{9}{2}.$ A unit cube with lower corner $(i,j,k),$ where $i,j,k\in\{0,1,2\},$ contains values of $x+y+z$ from $s=i+j+k$ to $s+3.$ It intersects the plane exactly when $s=2,$ $s=3,$ or $s=4.$ The coefficients of $x^2,$ $x^3,$ $x^4$ in $$(1+x+x^2)^3$$ are $6,$ $7,$ $6.$ Thus the plane intersects $6+7+6=19$ unit cubes, and the correct answer is D.