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As per Stokes drag formula, the terminal velocity is proportional to  square of radius.  so it is  (d) 1 : 4
There is also another method for terminal velocity in a fluid.  But it gives a different result.

     Here we can apply the equations of dynamics involving  gravity force, drag force due to the air , and the buoyancy force due to air.  We assume that the densities of both hail stones are same.  The hail stones are falling from a large height.  So we can assume that they reach some sort of terminal velocity.  That is the velocity remains same.  The gravitational force is equal to the drag force and buoyancy in opposite directions.

V = volume of stones = 4/3 * πR³
d = relative density of hail stone  wrt water
d_a = density of air layer  near the surface of Earth
v = terminal velocity of a solid travelling in a viscous fluid
μ = viscosity constant
Cd = drag coefficient
R = radius of the solid falling
A = surface area of object, that is projected on a plane = πR²

     Fg = m g = gravity = d * 4/3π R³ * g
     Fb = buoyancy = d_a * 4/3 π R³ * g

1)  Stokes law and  stokes drag
       Fd = drag force = 6 π μ R v      :  this is for Viscous flow of  a fluid around a solid spherical body.

   So  Fd = Fb + Fg      =>    v = 2/9 (d - d_w) g R² / μ

2)  A formula is based on the formulation that the drag in air  in the atmosphere is
             Fd = 1/2 * Cd * d_a * A * v²  = 1/2 Cd  d_a  πR² v²
   Then we get :    v² = (8π/3) R (d/d_a - 1) / Cd

       Here in this case,  v is proportional to the square root of Radius.
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