The correct answer is:
$ \displaystyle \text{84 c}{{\text{m}}^{2}}$

Explanation:
The semi-perimeter $ \displaystyle s=\dfrac{{13+14+15}}{2}=21\text{ cm}$

Using Heron’s formula:
Area $ \displaystyle \begin{array}{l}=\sqrt{{s\left( {s-a} \right)\left( {s-b} \right)\left( {s-c} \right)}}=\sqrt{{21\left( {21-13} \right)\left( {21-14} \right)\left( {21-15} \right)}}\\=\sqrt{{21\times 8\times 7\times 6}}=\sqrt{{7056}}=84\text{ c}{{\text{m}}^{2}}\end{array}$