You've to minimize the distance between (x, x^2/4) and (0, b) where b can take different values greater than 0.
Take b=2 and find the distance. Then take b=3 and find the distance. The distance has increased in case of b=3 so the answer has to be b<2.
But this method took time so I'm asking for any other method that is faster.
Set of all points equidistant from (0,b) is a circle. So set of all points at a distance of b from this point is a circle around it touching the x axis (tangent to it as well). At what b will this circle intersect with the parabola?
Solve x^2+(y-b)^2=b^2 with y=x^2/4
y=0, y=2b-4, obviously for the root to make sense it has to have a positive y, since both the circle and parabola lie on y.