Correct Answer: $54$

Solution :
In $\Delta \mathrm{ABC}$,
Given that $\angle \mathrm{BAE} = 30^{\circ}$
⇒ $\angle \mathrm{DAE = \angle BAE} = 30^{\circ}$  ($\mathrm{AE}$ is the bisector of $\angle \mathrm{BAD}$)
Now, $\angle \mathrm{BAD = \angle BAE + \angle DAE}$
⇒ $\angle \mathrm{BAD} = 30^{\circ} + 30^{\circ}$
⇒ $\angle \mathrm{BAD} = 60^{\circ}$
Also, $\angle \mathrm{CAD = \angle BAD} = 60^{\circ}$  ( $ \mathrm{AD}$ is the bisector of $\angle \mathrm{BAC}$)
⇒ $\angle \mathrm{CAE = \angle CAD + \angle DAE} = 60^{\circ} + 30^{\circ}$
In $\Delta \mathrm{CAE}$,
⇒ $\angle \mathrm{ CAE }= 90^{\circ}$
$\angle \mathrm{CAE} = 90^{\circ}$
$ \mathrm{AE} = 9 \;\mathrm{cm}$
$ \mathrm{EC} = 15 \;\mathrm{cm}$
⇒ $ \mathrm{AC} = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = 12\;\mathrm{cm}$
So, the area of $\Delta \mathrm{CAE}=\frac{1}{2} \times \mathrm{AE \times AC} = \frac{1}{2} \times 9 \times 12 = 54\;\mathrm{cm^2}$
Hence, the correct answer is $54$.