6. In an $8 \times 8$ square chessboard composed of 64 small squares, a circle with a radius of 4 is placed. If the sum of the areas inside the circle for all small squares that the circle's circumference passes through is denoted as $S_{1}$, and the sum of the areas outside the circle for all small squares that the circle's circumference passes through is denoted as $S_{2}$, then the integer part of $\frac{S_{1}}{S_{2}}$ is
(A) 0 ;
(B) 1 ;
(C) 2 ;
(D) 3 .
6. (B)

Solution According to the symmetry of the square, we only need to consider its $\frac{1}{4}$ part. Let the sum of the areas inside the circle of all small squares that the circumference passes through be $S_{1}^{\prime}$, and the sum of the areas outside the circle of all small squares that the circumference passes through be $S^{\prime}{ }_{2}$, then
$$
\begin{array}{l} 
S^{\prime}{ }_{1}=4 \pi-8, \quad S^{\prime}{ }_{2}=15-4 \pi . \\
\therefore \quad \frac{S_{1}}{S_{2}}=\frac{4 S_{1}^{\prime}}{4 S_{1}^{\prime}}=\frac{4 \pi-8}{15-4 \pi} \approx \frac{4.56}{2.44} . \\
S_{1}
\end{array}
$$

Therefore, the integer part of $\frac{S_{1}}{S_{2}}$ is 1.