According  to the question we have,

The radius of the Cylindrical tub (r) $=5cm$

Height of the Cylindrical tub (H) $=9.8cm$

Height of the cone outside the hemisphere (h) $=5cm$

Radius of the hemisphere $=3.5cm$

Then, we all know that

The volume of the Cylindrical tub $\left( {{V}_{1}} \right)=\pi {{r}^{2}}H$

${{V}_{1}}=\pi {{\left( 5 \right)}^{2}}=9.8$

${{V}_{1}}=770c{{m}^{3}}$

Then, the volume of the Hemisphere $\left( {{V}_{2}} \right)=2/3\times \pi \times {{r}^{3}}$

${{V}_{2}}=2/3\times 22/7\times {{3.5}^{3}}$

${{V}_{2}}=89.79c{{m}^{3}}$

And, the volume of the Hemisphere $\left( {{V}_{3}} \right)=23\times \pi \times r\times 2h$

${{V}_{3}}=2/3\times 22/7\times {{3.5}^{2}}\times 5$

${{V}_{3}}=64.14c{{m}^{3}}$

Therefore, total volume (V) $=$ Volume of the cone $+$ Volume of the hemisphere

$={{V}_{2}}+{{V}_{3}}$

$V=89.79+64.14c{{m}^{3}}$

$=154c{{m}^{3}}$

Thus, the total volume of the solid $=154c{{m}^{3}}$

In order to find the volume of the water left in the tube,  subtract the volume of the hemisphere and the cone from the volume of the cylinder.

Therefore, the volume of water left in the tube $={{V}_{1}}-{{V}_{2}}$

$=770–154$

$=616c{{m}^{3}}$

Hence, the volume of water left in the tube is $616c{{m}^{3}}$.