Question 529102
As stated, there is no single answer to that problem.
I suppose what was meant was that rectangle ABCD is similar to the 3 small congruent rectangles. That makes the problem solvable. 
Since I have not figured out how to draw on this website yet, I'll ned to use your imagination.
Visualize points A, E, F, and B equally spaced on a horizontal line.
directly below then, imagine points D, G, H, and C.
Rectangle ABCD is similar to rectangles AEGD, EFHG, and FBCH, meaning that the ratios of length to with are the same.
The distances AE, EF, FB, DG, GH, and HC are all 3, the width of the small rectangles. They add up to a length of 9 for the large rectangle.
The length of the small rectangles is the distance AD, the width of the large rectangle. We'll call ix {{{x}}}.
The length to width ratio is
{{{9/x}}} for the large rectangle and
{{{x/3}}} for the small rectangles, so
{{{x/3=9/x}}} and {{{x=sqrt(27)=3sqrt(3)}}}
making the area of the large rectangle
{{{9(3sqrt(3))=27sqrt(3)}}}