Let $h$ be the depth at which the rubber ball is taken. Then
$p=h\rho g \quad\ldots\left(i\right)$
By definition of bulk modulus,
$B=-\frac{P}{\Delta V /V}$
The negative sign shows that with increase in pressure, a decrease in volume occurs.
$\therefore\quad$ $P=B \frac{\Delta V}{V}$
Using $\left(i\right)$,
$h\rho g$=$B \frac{\Delta V}{V}$ $\quad$ or $\quad$ $h=\frac{B}{\rho g}$ $\frac{\Delta V}{V}$
Substituting the given values, we get
$h=\frac{9\times10^{8} N\,m^{-2}}{10^{3}kg \,m^{-3}\times10 \,ms^{-2}}$ $\left(\frac{0.1}{100}\right)$ $=90\, m\quad$