So, $w_{A}=m g$

$Case$ $B$ In this case, downward forces are weight and reaction of buoyant force.

$\therefore \quad w_{B}=m g+F_{B}$

$Case$ $C$ In this case, downward acting forces are weight and reaction of buoyant force which is same is as that of case $B$ as balls are of same size.

$\therefore \quad w_{C}=m g+F_{B}$

$Case$ $D$ ln this case, forces acting on bottom of beaker are

$m^{\prime} g$ : weight of ball

$m g$ : weight of water

$F_{B}$ : reaction of buoyant force

$T$ : tension in string

Also, $\quad T=F_{B}$

So, net downward force on bottom of beaker is $m g+m^{\prime} g$.

$\therefore \quad w_{D}=m g+m^{\prime} g$

So, $w_{B}=w_{C} > w_{D} > w_{A}$