$\begin{array}{l}\mathrm{\Delta P}=\mathrm{B}\left(\frac{\mathrm{V}}{\mathrm{V}}\right)\text{ Here }\left(\frac{\mathrm{V}}{\mathrm{V}}\right)=\frac{0\cdot 1}{100}\\ \therefore \mathrm{\Delta P}=9\times {10}^5\end{array}$
Atmospheric pressure at sea level =10 m of water column.
Let the ball is taken to a depth h. Then
$\begin{array}{l}\mathrm{\Delta P}=(\mathrm{h}-10)\mathrm{\rho g}\\ 9\times {10}^5=(\mathrm{h}-10)\left(1000\right)\\ \therefore \mathrm{h}=100\mathrm{m}\end{array}$