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.