Let V be the volume of ball
The ball attains a velocity
$v =\sqrt {2gh}$
$\quad= \sqrt {2 \times 9.8 \times 10}$
$\quad= \sqrt {196}\; m/s$
Inside water - density of ball $= 0.5 \times 10^3\;kg/m^3$
density of water $=1 \times 10^3\;kg/m^3$
Net retardation produced
$ \qquad=\large\frac{upthrust - weight}{mass}$
$\qquad =\large\frac{V \times 10^3 \times g - V \times 0.5 \times 10^3 \times g}{V \times 0.5 \times 10^3}$
$\qquad=g$
Since the effective retardation inside liquid $=g$ the ball will fall $10\; m$ inside water before raising up.
Hence c is the correct answer.