Six cubes, each an inch on an edge, are fastened together, as shown.  Find the total surface area in square inches.  Include the top, bottom and sides.

[asy]/* AMC8 2002 #22 Problem */
draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);
draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1,1));
draw((1,0)--(1.5,0.5)--(1.5,1.5));
draw((0.5,1.5)--(1,2)--(1.5,2));
draw((1.5,1.5)--(1.5,3.5)--(2,4)--(3,4)--(2.5,3.5)--(2.5,0.5)--(1.5,.5));
draw((1.5,3.5)--(2.5,3.5));
draw((1.5,1.5)--(3.5,1.5)--(3.5,2.5)--(1.5,2.5));
draw((3,4)--(3,3)--(2.5,2.5));
draw((3,3)--(4,3)--(4,2)--(3.5,1.5));
draw((4,3)--(3.5,2.5));
draw((2.5,.5)--(3,1)--(3,1.5));[/asy]

$ \textbf{(A)}\ 18\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 36$
1. Each cube has 6 faces, and since there are 6 cubes, the total surface area if they were not connected would be:
   \[
   6 \times 6 = 36 \text{ square inches}
   \]

2. When cubes are connected, the faces that are in contact do not contribute to the surface area. We need to count the number of such faces.

3. Let's analyze the connections:
   - The front cube connects with 1 other cube.
   - The bottom-left-back cube connects with 2 other cubes.
   - The bottom-middle cube connects with 2 other cubes.
   - The second story-leftmost cube connects with 3 other cubes.
   - The second-story-rightmost cube connects with 1 other cube.
   - The top cube connects with 1 other cube.

4. Each connection between two cubes hides 2 faces (one from each cube). Therefore, we need to count the total number of connections:
   \[
   1 + 2 + 2 + 3 + 1 + 1 = 10 \text{ hidden faces}
   \]

5. Subtract the number of hidden faces from the total surface area:
   \[
   36 - 10 = 26 \text{ square inches}
   \]

Conclusion:
\[
\boxed{26}
\]