Solution:

Find projections of all given circles on one of the sides of the square. The projection of each circle is a segment whose length is equal to the length of a diameter of this circle. Since the sum of the lengths of all circles' diameters is equal to $10 / \pi$, it follows that the sum of the lengths of all mentioned projections is equal to $10 / \pi > 3$. Because the side of the square is equal to $1$, we conclude that at least one point is covered with at least four of these projections. Hence, a perpendicular line to the projection side passing through this point meets at least four of the given circles, so this is a line with the desired property.