Answer
$0.0442\;kg$
Work Step by Step
The forces acting on the ball in upward direction are the normal force $(F_n)$ and the buoyant force $(F_b)$
$F_u=F_n+F_b$
The force acting on the ball in downward direction is the weight of the ball $(W)$
$F_d=W$
At equllibrium
$F_d=F_u$
or, $W=F_n+F_b$
or, $mg=F_n+\rho gV_{sub}$
or, $m=\frac{F_n+\rho g\frac{4\pi r^3}{3}}{g}$
Substituting the given values
$m=\frac{9.48\times10^{-2}+1030 \times9.81\times \frac{4\times\pi \times 0.02^3}{3}}{9.81}\;kg$
or, $m=0.0442\;kg$
Therefore, the mass of the ball $0.0442\;kg$
