Holding a rectangular sheet of paper $ABCD$, Prair folds triangle $ABD$ over diagonal $BD$, so that the new location of point $A$ is $A'$. She notices that $A'C =\frac13 BD$. If the area of $ABCD$ is $27\sqrt2$, find $BD$.
1. Let \( E \) be the intersection of \( A'D \) and \( BC \), and let \( AB = x \) and \( AD = y \). Since the area of \( ABCD \) is \( 27\sqrt{2} \), we have:
   \[
   xy = 27\sqrt{2}
   \]

2. Since \( \triangle A'EC \sim \triangle DEB \), we have:
   \[
   \frac{ED}{EA'} = \frac{A'C}{BD} = \frac{1}{3}
   \]
   Given that \( A'C = \frac{1}{3} BD \), and since reflection preserves lengths, we have:
   \[
   A'D = AD = y
   \]

3. From the similarity ratio, we know:
   \[
   ED = \frac{3}{4} y \quad \text{and} \quad EA' = \frac{1}{4} y
   \]

4. Using the Pythagorean Theorem in \( \triangle A'EC \):
   \[
   EC^2 + CD^2 = ED^2
   \]
   Substituting the known lengths:
   \[
   \left(\frac{1}{4} y\right)^2 + x^2 = \left(\frac{3}{4} y\right)^2
   \]
   Simplifying:
   \[
   \frac{1}{16} y^2 + x^2 = \frac{9}{16} y^2
   \]
   Solving for \( x^2 \):
   \[
   x^2 = \frac{9}{16} y^2 - \frac{1}{16} y^2 = \frac{8}{16} y^2 = \frac{1}{2} y^2
   \]
   Therefore:
   \[
   y^2 = 2 x^2 \quad \text{and} \quad y = \sqrt{2} x
   \]

5. Substituting \( y = \sqrt{2} x \) into \( xy = 27\sqrt{2} \):
   \[
   x (\sqrt{2} x) = 27\sqrt{2}
   \]
   Simplifying:
   \[
   \sqrt{2} x^2 = 27\sqrt{2}
   \]
   Solving for \( x^2 \):
   \[
   x^2 = 27 \quad \text{and} \quad x = 3\sqrt{3}
   \]
   Therefore:
   \[
   y = \sqrt{2} x = \sqrt{2} \cdot 3\sqrt{3} = 3\sqrt{6}
   \]

6. Finally, using the Pythagorean Theorem to find \( BD \):
   \[
   BD = \sqrt{AB^2 + AD^2} = \sqrt{x^2 + y^2}
   \]
   Substituting the values of \( x \) and \( y \):
   \[
   BD = \sqrt{(3\sqrt{3})^2 + (3\sqrt{6})^2} = \sqrt{27 + 54} = \sqrt{81} = 9
   \]

The final answer is \(\boxed{9}\)