Think of the magnetic field as a broadcast signal; just as a light source appears dimmer further away, the 'density' of the magnetic field at the center decreases as the size of the coil loop increases.

According to the Biot-Savart Law, the magnetic field $B$ at the center of a circular coil of $N$ turns, radius $r$, and carrying current $I$ is given by $B = \frac{\mu_{0} N I}{2r}$. Since $I$ and $N$ are identical for both coils, the relationship simplifies to $B \propto \frac{1}{r}$. Therefore, doubling the radius $r$ results in the magnetic field being halved.

Always verify if the question asks for the center of the coil or a point on the axis, as the axial formula $B = \frac{\mu_{0} N I R^2}{2(R^2 + x^2)^{3/2}}$ behaves differently.