As per the given data,

Perimeter of the upper end of  a frustum of a right circular cone $=44cm$

So, $2\pi {{r}_{1}}=44$

$2\left( 22/7 \right){{r}_{1}}=44$

(radius of upper end of a frustum of a right circular cone ) ${{r}_{1}}=7cm$

Perimeter of the lower end of a frustum of a right circular cone $=33cm$

So, $2\pi {{r}_{2}}=33$

$2\left( 22/7 \right){{r}_{2}}=33$

${{r}_{2}}=21/4cm$

Now,

Let us assume the slant height of the frustum of a right circular cone be L.

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

$=\sqrt{{{\left( 7-5.25 \right)}^{2}}}+{{16}^{2}}$

Therefore, L $=16.1cm$

As we known that,

the curved surface area of the frustum cone  is given by formula $\pi \left( {{r}_{1}}+{{r}_{2}} \right)l$

$=\pi \left( 7+5.25 \right)16.1$

Therefore, curved surface area of the frustum cone is $619.65c{{m}^{2}}$

Next, according to the question,

The volume of the frustum cone is given by the formula $1/3\pi \left( r_{2}^{2}+r_{1}^{2}+{{r}_{1}}{{r}_{2}} \right)h$

By substituting the given values,

$=1/3\pi \left( {{7}^{2}}+{{5.25}^{2}}+\left( 7 \right)\left( 5.25 \right) \right)\times 16$

$=1898.56c{{m}^{3}}$

Thus, volume of the cone is $1898.56c{{m}^{3}}$

As we known that  the total surface area of the frustum cone

$=\pi \left( {{r}_{1}}+{{r}_{2}} \right)\times L+\pi r_{1}^{2}+\pi r_{2}^{2}$

So by putting given values,

$=\pi \left( 7+5.25 \right)\times 16.1+\pi {{7}^{2}}+\pi {{5.25}^{2}}$

$=\pi \left( 7+5.25 \right)\times 16.1+\pi \left( {{7}^{2}}+{{5.25}^{2}} \right)=860.27c{{m}^{2}}$

Therefore, the total surface area of the frustum cone is equal  $860.27c{{m}^{2}}$