According to the question,

Radius of the common base (r) $=3.5cm$

Height of the cylindrical part (h) $=10cm$

Height of the conical part (H) $=6cm$

Assume, ‘l’ be the slant height of the cone

Now, we all know that

${{l}^{2}}={{r}^{2}}+{{H}^{2}}$

${{l}^{2}}={{3.5}^{2}}+{{6}^{2}}=12.25+36=48.25$

$l=6.95cm$

Thus, the curved surface area of the cone $\left( {{S}_{1}} \right)=\pi rl$

${{S}_{1}}=\pi \left( 3.5 \right)\left( 6.95 \right)$

${{S}_{1}}=76.38c{{m}^{2}}$

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

${{S}_{2}}=2\pi {{\left( 3.5 \right)}^{2}}$

${{S}_{2}}=77c{{m}^{2}}$

Now, the curved surface area of the cylinder $\left( {{S}_{3}} \right)=2\pi rh$

${{S}_{2}}=2\pi \left( 3.5 \right)\left( 10 \right)$

${{S}_{2}}=220c{{m}^{2}}$

So, the total surface area $\left( S \right)={{S}_{1}}+{{S}_{2}}+{{S}_{3}}$

$S=76.38+77+220=373.38c{{m}^{2}}$

Hence, the total surface area of the solid is $373.38c{{m}^{2}}$