According to the question it is given that,

Radius of the hemispherical end (r) $=7cm$

Height of the solid $=(h+2r)=104cm$

$\Rightarrow h+2r=104$

$\Rightarrow h=104-\left( 2\times 7 \right)$

Then, $h=90cm$

As we all know that,

The curved surface area of the cylinder $\left( {{S}_{1}} \right)=2\pi rh$

${{S}_{1}}=2\pi \left( 7 \right)\left( 90 \right)$

${{S}_{1}}=3960c{{m}^{2}}$

Now, curved surface area of the two hemisphere $\left( {{S}_{2}} \right)=2\left( 2\pi {{r}^{2}} \right)$

${{S}_{2}}=2\times 2\pi {{\left( 7 \right)}^{2}}$

${{S}_{2}}=616c{{m}^{2}}$

Thus,

The total curved surface area of the solid (S) $=$ Curved surface area of the cylinder $+$ Curved surface area of the two hemispheres

$S={{S}_{1}}+{{S}_{2}}$

$S=3960+616$

$S=4576c{{m}^{2}}=45.76d{{m}^{2}}$

It is given that the cost of polishing the  surface of the solid is $Rs.10$

Thus, the cost of polishing the $45.76d{{m}^{2}}$ surface of the solid$=Rs\left( 10\times 45.76 \right)=rs.457.6$

Hence,

The cost of polishing the whole surface of the solid is $Rs.457.60$