Let the radius of the base and slant height of the cone be $$r\ cm$$ and $$I\ cm$$, respectively
Height of the cone, $$h=24\ cm$$
Volume of the cone $$=1212\ cm^3$$
$$\therefore \pi r^2 h=1212\ cm^3$$
$$\Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times r^2 \times 24=1212$$
$$\Rightarrow r^2=\dfrac{1212 \times 7}{22 \times 8}=48.2$$
$$\Rightarrow r=7\ cm$$ (Approx)
Now,
$$I^2=r^2+h^2$$
$$\Rightarrow I^2=(7)^2+(24)^2$$
$$\Rightarrow I^2=49+573=625$$
$$\Rightarrow I=\sqrt{625}=25\ cm$$ (Approx)
$$\therefore$$ Surface area of the cone $$=\pi r I=\dfrac{22}{7} \times 7 \times 25=550\ cm^2$$ (approx)
Thus, the surface area of the cone is approximately $$500\ cm^2$$