Let the old centripetal force be "F_c = \\frac{m v^2}{r}". When the radius is doubled, the new centripetal force is "F_c' = \\frac{m v^2}{2 r}". Dividing the new force by the old, obtain "\\frac{F_c'}{F_c} = \\frac{r}{2 r} = \\frac{1}{2}", so "F_c' = \\frac{F_c}{2}" - the centripetal force is halved.
The answer is 4).
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