Solution:

WLOG, let the width of the rectangle be $1$ and the length be $\sqrt3.$

Draw the line perpendicular to the midpoint of the fold, as shown below.

Note that $QR = \dfrac{2\sqrt3}{3}$ and $$QT = \sqrt{1^2 + \left(\dfrac{\sqrt3}{3}\right)^2}$$$$= \sqrt{1 + \dfrac{1}{3}} = \sqrt{\dfrac{4}{3}}.$$ This tells us $$QT = \dfrac{2\sqrt3}{3} = QR.$$ This means that \triangleriangle QRT is equilateral. Similarly, \triangleriangle RTS is equilateral. This makes the two triangles congruent.

This means that after the rectangle gets folded, this area will be overlapped. The area of the rectangle is $1 \cdot \sqrt3 = \sqrt3.$ The side length of this triangle is $QT = \dfrac{2\sqrt3}{3}.$ The area of it is then $$\left(\dfrac{2\sqrt3}{3}\right)^2 \cdot \dfrac{\sqrt3}{4} = \dfrac{\sqrt3}{3}.$$ The area of the folded figure is then $$\sqrt3 - \dfrac{\sqrt3}{3} = \dfrac{2\sqrt3}{3}.$$ The desired ratio is then $$\dfrac{2\sqrt3}{3} \div \sqrt3 = \dfrac{2}{3}.$$ Thus, C is the correct answer.