we know that
the solid of the figure is a sphere
the volume of the sphere is equal to
[tex]V=\frac{4}{3}\cdot\pi\cdot r^3[/tex]we have
pi=3.14
r=25 m
substitute
[tex]\begin{gathered} V=\frac{4}{3}\cdot3.14\cdot25^3 \\ V=65,416.67\text{ m\textasciicircum{}3} \end{gathered}[/tex]