6. (3 points) Two differently sized cubic building blocks are glued together, forming the solid figure shown in the diagram. The four vertices of the smaller cube's glued face are the non-midpoint quarter points of the larger cube's glued face edges. If the edge length of the larger cube is 4, then the surface area of this solid figure is $\qquad$ .


【Analysis】From the figure, we can see that the two connected faces of the large cube and the small cube are as shown: Since the edge length of the large cube is 4, the legs of the right-angled triangles on all four sides are $3$ and $1$; if the four right-angled triangles on all four sides are cut out, they can be exactly pieced together to form a square with an area of $4 \times 4 - 3 \times 1 \div 2 \times 4 = 10$. According to the area formula of a square, the area of one face of the large cube is $4 \times 4 = 16$; thus, the surface area of this solid is the sum of the surface area of the large cube and the areas of the four faces of the small cube, which can be solved accordingly.

【Solution】According to the analysis of the problem, we can get: the area of one face of the large cube is: $4 \times 4 = 16$, the area of one face of the small cube is:
$$
\begin{array}{l}
4 \times 4 - 3 \times 1 \div 2 \times 4 \\
= 16 - 6 \\
= 10,
\end{array}
$$

Therefore, the surface area of this solid figure is:
$$
\begin{array}{l}
16 \times 6 + 10 \times 4 \\
= 96 + 40 \\
= 136,
\end{array}
$$

Answer: The surface area of this solid figure is 136.
So the answer is: 136.