The large cube has dimensions $3\times 3\times 3 = 27$ small cubes.
The cubes with red paint on exactly one face are the ones on the surface of the large cube.
There are 8 cubes on the corners, and each of them has 3 faces painted red.
There are 12 cubes on the edges, and each of them has 2 faces painted red.
There are 6 cubes on the faces that are not on the edges, and each of them has 1 face painted red.
So, there are $8\times 3 + 12\times 2 + 6\times 1 = 24 + 24 + 6 = 54$ cubes with red paint on exactly one face.
The cubes with red paint on at least two faces are the ones on the edges and corners of the large cube.
There are 8 cubes on the corners and 12 cubes on the edges, so there are $8 + 12 = \boxed{20}$ cubes with red paint on at least two faces.
The answer is: 20
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
The six faces of a three-inch wooden cube are each painted red. The cube is then cut into one-inch cubes along the lines shown in the diagram. How many of the one-inch cubes have red paint on at least two faces? [asy]

pair A,B,C,D,E,F,G;

pair a,c,d,f,g,i,j,l,m,o,p,r,s,u,v,x,b,h;

A=(0.8,1);
B=(0,1.2);
C=(1.6,1.3);
D=(0.8,0);
E=B-(A-D);
F=C-(A-D);
G=B+(C-A);

draw(E--D--F--C--G--B--A--D);
draw(A--C); draw(E--B);

a=(1/3)*D+(2/3)*E;
c=(2/3)*D+(1/3)*E;
p=(1/3)*A+(2/3)*B;
r=(2/3)*A+(1/3)*B;

draw(a--p);
draw(c--r);

v=(1/3)*B+(2/3)*E;
x=(2/3)*B+(1/3)*E;
b=(1/3)*A+(2/3)*D;
h=(2/3)*A+(1/3)*D;

draw(v--b);
draw(x--h);

s=(1/3)*C+(2/3)*A;
u=(2/3)*C+(1/3)*A;
d=(1/3)*F+(2/3)*D;
f=(2/3)*F+(1/3)*D;

draw(s--d);
draw(f--u);

g=(1/3)*C+(2/3)*F;
i=(2/3)*C+(1/3)*F;

draw(i--h);
draw(g--b);

m=(1/3)*B+(2/3)*G;
o=(2/3)*B+(1/3)*G;

draw(m--u);
draw(o--s);

j=(1/3)*G+(2/3)*C;
l=(2/3)*G+(1/3)*C;

draw(l--p);
draw(r--j);

[/asy]

### Answer:
