In a triangle $\mathrm{ABC}, \mathrm{AD}$ and $\mathrm{BE}$ are medians. If $\mathrm{AD}=4$, $\left\lfloor\mathrm{DAB}=\frac{\pi}{6}\right.$ and $\left\lfloor\mathrm{ABE}=\frac{\pi}{3}\right.$ then the area of $\triangle \mathrm{ABC}$ is

• A

  $\frac{14}{3 \sqrt{3}}$
• B

  $\frac{28}{3 \sqrt{3}}$
• C

  $\frac{11}{3 \sqrt{3}}$
• Correct answer
  D

  $\frac{32}{3 \sqrt{3}}$

Hence, the correct answer is option: 'D'.