Given:
- There is 1 large circle with diameter $D$.
- Inside it, there are 7 smaller circles of equal diameter arranged as shown.
- The smaller circles are shaded; the unshaded regions are the parts inside the large circle but outside the small circles.
- We are asked to find the sum of the areas of the unshaded regions inside the large circle in terms of $D$.

Step 1: Understand the arrangement of the small circles
The figure shows 7 small circles packed inside one large circle:
- One small circle in the center.
- Six small circles arranged around it, touching the center small circle and touching adjacent small circles.

This is a classic circle packing arrangement known as "one circle surrounded by six circles of equal size," all inside the large circle.

Step 2: Relationship between the small circles and large circle diameters
- Let the diameter of each small circle be $d$.
- Since there are 7 small circles arranged in a hexagonal pattern with 6 around the center, the large circle exactly encloses them all, touching the outer small circles.
- The radius of the large circle $\(R=\frac{D}{2}\)$.
- The centers of the 6 outer small circles lie on a circle around the center small circle.

Because each small circle's diameter is $d$, radius is $\(r=\frac{d}{2}\)$.
The distance from the center of the big circle to the center of any outer small circle is:

$$
\(r+d=r+2 r=3 r\)
$$

(because the outer circles touch the center circle, distance between centers is $\(d=2 r\)$, and the radius of the large circle must cover the center circle radius $r$ plus the radius $r$ of the outer circle).

But to touch the large circle boundary, the radius $R$ should equal:

$$
\(R=r+2 r=3 r\)
$$


Thus,

$$
\(R=3 r \Longrightarrow r=\frac{R}{3}\)
$$

and since $\(r=\frac{d}{2}\)$,

$$
\(d=2 r=\frac{2 R}{3}=\frac{2}{3} \times \frac{D}{2}=\frac{D}{3}\)
$$

So,
$\(d=\frac{D}{3} \quad\)$ (diameter of each small circle)

Step 3: Calculate the area of the large circle

$$
\(A_{\text {large }}=\pi R^2=\pi\left(\frac{D}{2}\right)^2=\pi \frac{D^2}{4}\)
$$


Step 4: Calculate the area of all 7 small circles
Each small circle has radius

$$
\(r=\frac{d}{2}=\frac{D}{6}\)
$$


Area of one small circle:

$$
\(A_{\text {small }}=\pi r^2=\pi\left(\frac{D}{6}\right)^2=\pi \frac{D^2}{36}\)
$$


Area of 7 small circles:

$$
\(7 \times A_{\text {small }}=7 \times \pi \frac{D^2}{36}=\frac{7 \pi D^2}{36}\)
$$

Step 5: Calculate the sum of unshaded areas inside the large circle
Unshaded regions inside large circle $=$ area of large circle - area of 7 small circles

$$
\(A_{\text {unshaded }}=A_{\text {large }}-7 A_{\text {small }}=\pi \frac{D^2}{4}-\frac{7 \pi D^2}{36}\)
$$


Find common denominator:

$$
\(\frac{D^2 \pi}{4}=\frac{9 \pi D^2}{36}\)
$$


Subtract:

$$
\(A_{\text {unshaded }}=\frac{9 \pi D^2}{36}-\frac{7 \pi D^2}{36}=\frac{2 \pi D^2}{36}=\frac{\pi D^2}{18}\)
$$


Final answer:
The sum of the areas of the unshaded regions inside the large circle is

$$
\(\frac{\pi D^2}{18}\)
$$


This corresponds to option:
B. $\(\pi D^2 / 18\)$