Correct Answer: $12$ cm

Solution :

Given: The base of a right pyramid is an equilateral triangle.
Side of a triangle $AB = 10\sqrt3$ cm
The total surface area of a pyramid $= 270\sqrt3$ cm ²
$\therefore$ Inradius of a triangle $OE = \frac{\text{side of equilateral triangle}}{2\sqrt3}$
= $\frac{10\sqrt3}{2\sqrt3}=5$ cm
Now, slant height (l) $= \sqrt {h^{2}+OE^{2}}$
$= \sqrt {h^{2}+25}$
Base area $=\frac{\sqrt{3}}{4}×(10\sqrt{3})^{2}$
Total surface area $= \frac{1}{2}×(\text{perimeter of the base×slant height)+base area}$
⇒ $270\sqrt3=\frac{1}{2}(10\sqrt{3}×3×\sqrt {h^{2}+25})+\frac{\sqrt{3}}{4}×(10\sqrt{3})^{2}$
⇒ $270\sqrt3=15\sqrt{3}(\sqrt{{h^{2}+25}})+75\sqrt{3}$
⇒ $15\sqrt{3}(\sqrt{{h^{2}+25}})=195\sqrt3$
⇒ $\sqrt{{h^{2}+25}}=13$
⇒ $h^{2}=144$
⇒ $h = 12$ cm
Hence, the correct answer is $12$ cm.