According to the given data in the question,

Height of frustum (h) $=8m$ (given)

Bigger and smaller radii of the frustum cone are $13cm$ and $7cm$.

Therefore, ${{r}_{1}}=13cm$ and ${{r}_{2}}=7cm$

Let us assume ‘L’ be slant height of the frustum cone.

Then, we know that the slant height is given by formula,

$L=\sqrt{{{\left( {{r}_{1}}-{{r}_{2}} \right)}^{2}}}+{{h}^{2}}$

$L=\sqrt{36+64}$

$L=10cm$

Therefore, Curved surface area of the frustum $\left( {{s}_{1}} \right)$ is given by $\pi \left( {{r}_{1}}+{{r}_{2}} \right)\times L$

$=\pi \left( 13+7 \right)\times 10$

$=200\pi {{m}^{2}}$

Thus, the given slant height of conical cap in the question is $12m$

Base radius of upper cap cone  given in question is  $7m$

So therefore, the curved surface area of upper cap cone $\left( {{s}_{2}} \right)=\pi rl$

$=\pi \times 7\times 12$

$=264{{m}^{2}}$

Thus, the total canvas required for making tent (S) is equal to ${{s}_{1}}+{{s}_{2}}$, which is

$s=200\pi +264=892.57{{m}^{2}}$

Therefore, $892.57{{m}^{2}}$of the canvas is required for making the tent.