Solution:

We know that: $$\begin{align*}[A'B'C'] =& [ABC]+ [A'B'A]\\ &+ [B'C'B] + [C'A'C] .\end{align*}$$ The last three terms on the right hand side of the equation have the same area, so the area: $$[A'B'C']=[ABC]+3[A'B'A].$$ Therefore, to find the ratio in question, we need to find: $$\dfrac{[ABC]+3[A'B'A] }{[ABC]}$$$$= 1+ 3\dfrac{[A'B'B]}{[ABC]}.$$ Then, $$\begin{align*}[A'B'B] =& \frac12 A'A\cdot A'A \\&\cdot \sin A'AB' \end{align*}$$ and $$[ABC] = \frac 12 AB\cdot AC \cdot \sin BAC.$$ Since $\angle A'AB'$ and $\angle BAC$ are supplements, they have the same sine.

Therefore, $$\dfrac{[A'B'B]}{[ABC]} = \dfrac{A'A \cdot B'A}{AB\cdot AC} .$$ Then, $$A'B = AB + BB' = 4AB,$$ and $$AA' = 3 AC .$$ This makes $$\dfrac{[A'B'B]}{[ABC]} = 4\cdot 3 = 12.$$ As such, the final ratio is $$1+ 3\cdot 12 = 37.$$

Thus, the correct answer is E .