6. (10 points) Two smaller cubic building blocks are attached to two faces of a larger cubic building block, forming the solid figure shown in the diagram. Each smaller block is attached such that the four vertices of its attaching face are one of the five equal division points on each side of the larger block's attaching face. If the edge lengths of the three blocks are all different and the largest edge length is 5, then the surface area of this solid figure is


$\underline{270}$.
【Analysis】Observing the figure, we can see that the two connected faces of the large cube and the small cubes are as shown in the figure: Since the edge length of the large cube is 5, the legs of the right-angled triangles on the four sides are $4$ and $1$ respectively; if the four right-angled triangles on the four sides are cut out, they can be exactly pieced together to form a square with an area of $5 \times 5-4 \times 1 \div 2 \times 5=15$, according to the area formula of a square, the area of one face of the large cube is $5 \times 5=25$; then the surface area of this cuboid is the sum of the surface area of the large cube and the areas of four faces of each of the two small cubes, 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: $5 \times 5=25$
The area of one face of the small cube is:
$$
\begin{array}{l}
5 \times 5-4 \times 1 \div 2 \times 5 \\
=25-10 \\
=15
\end{array}
$$

Therefore, the surface area of this solid figure is:
$$
\begin{aligned}
& 25 \times 6+15 \times 4 \times 2 \\
& =150+120 \\
= & 270
\end{aligned}
$$

Answer: The surface area of this solid figure is 270.
So the answer is: 270 .