2. The mass of the first iron ball is $1462.5 \%$ greater than the mass of the second ball. By what percentage will less paint be needed to paint the second ball compared to the first? The volume of a sphere with radius $R$ is $\frac{4}{3} \pi R^{3}$, and the surface area of a sphere is $4 \pi R^{2}$.
2. Answer: $84 \%$.

Let's denote the radii of the spheres as $R$ and $r$ respectively. Then the first condition means that

$$
\frac{\frac{4}{3} \pi R^{3}-\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi r^{3}} \cdot 100=1462.5 \Leftrightarrow \frac{R^{3}-r^{3}}{r^{3}}=14.625 \Leftrightarrow \frac{R^{3}}{r^{3}}=\frac{125}{8} \Leftrightarrow R=\frac{5}{2} r . \text { The sought }
$$

percentage is: $\frac{4 \pi R^{2}-4 \pi r^{2}}{4 \pi R^{2}} \cdot 100=\frac{R^{2}-r^{2}}{R^{2}} \cdot 100=\left(1-\left(\frac{r}{R}\right)^{2}\right) \cdot 100$

$=\left(1-\left(\frac{2}{5}\right)^{2}\right) \cdot 100=\frac{2100}{25}=84$.