Correct Answer: $(16\sqrt{3}+48\sqrt{35})$

Solution :
Total Surface area of a pyramid = Base area + 3 × The area of each triangular face
The area of a equilateral triangle with sides 8 cm = $\frac{\sqrt{3}}{4}×(\text{side})^2=\frac{\sqrt{3}}{4}×8^2=16\sqrt{3}\ \text{cm}^2$
Let the height of the pyramid be $h$.
$h^2=\text{Slant height}^2-(\frac{1}{2}×\text{base})^2$
$⇒h^2=24^2-(\frac{1}{2}×8)^2$
$h^2=576-16$
$\therefore h=\sqrt{560}=4\sqrt{35}\ \text{cm}$
The area of each triangular face = $\frac{1}{2}×$ base × height = $\frac{1}{2}×8×4\sqrt{35}=16\sqrt{35}\ \text{cm}^2$
Total surface area of a pyramid = $16\sqrt{3}+3×16\sqrt{35}=(16\sqrt{3}+48\sqrt{35})\ \text{cm}^2$
Hence, the correct answer is $(16\sqrt{3}+48\sqrt{35})\ \text{cm}^2$.