Six regular hexagonal blocks of side length 1 unit are arranged inside a regular hexagonal frame. Each block lies along an inside edge of the frame and is aligned with two other blocks, as shown in the figure below. The distance from any corner of the frame to the nearest vertex of a block is $\frac{3}{7}$ unit. What is the area of the region inside the frame not occupied by the blocks?
[asy] unitsize(1cm); draw(scale(3)*polygon(6)); filldraw(shift(dir(0)*2+dir(120)*0.4)*polygon(6), lightgray); filldraw(shift(dir(60)*2+dir(180)*0.4)*polygon(6), lightgray); filldraw(shift(dir(120)*2+dir(240)*0.4)*polygon(6), lightgray); filldraw(shift(dir(180)*2+dir(300)*0.4)*polygon(6), lightgray); filldraw(shift(dir(240)*2+dir(360)*0.4)*polygon(6), lightgray); filldraw(shift(dir(300)*2+dir(420)*0.4)*polygon(6), lightgray); [/asy]
$\textbf{(A)}~\frac{13 \sqrt{3}}{3}\qquad\textbf{(B)}~\frac{216 \sqrt{3}}{49}\qquad\textbf{(C)}~\frac{9 \sqrt{3}}{2} \qquad\textbf{(D)}~ \frac{14 \sqrt{3}}{3}\qquad\textbf{(E)}~\frac{243 \sqrt{3}}{49}$
Examining the red isosceles trapezoid with $1$ and $\dfrac{3}{7}$ as two bases, we know that the side lengths are $\dfrac{4}{7}$ from $30-60-90$ triangle.
We can conclude that the big hexagon has side length 3.
Thus the target area is: area of the big hexagon - 6 * area of the small hexagon.
$\dfrac{3\sqrt{3}}{2}(3^2-6\cdot1^2) = \dfrac{3\sqrt{3}}{2}(3) = \boxed{\textbf{(C)}~\frac{9 \sqrt{3}}{2}}$
~Technodoggo

Solution 1.1
We can extend the line of the parallelogram to the end until it touches the next hexagon and it will make a small equilateral triangle and a longer parallelogram. We can prove that one side of the tiny equilateral triangle is 4/7 by playing around with angles and the parallelogram because it is parallel, we can then use the whole side of the hexagon which is one and subtract 3/7 which is one side of the equilateral triangle which is 4/7. That means the whole side of the big hexagon length is 3 and we can continue with solution 1.

-BrandonChen