Two circular coils of radii 5.0 cm and 10 cm carry equal currents...

The magnetic field at the centre of a circular coil of $N$ turns, radius $r$, and carrying current $I$ is given by $B = \dfrac{\mu_0 N I}{2r}$.

For the first coil:
$B_1 = \dfrac{\mu_0 N_1 I_1}{2r_1} = \dfrac{4\pi \times 10^{-7} \times 50 \times 2.0}{2 \times 0.05} = 4\pi \times 10^{-4} \text{ T}$

For the second coil:
$B_2 = \dfrac{\mu_0 N_2 I_2}{2r_2} = \dfrac{4\pi \times 10^{-7} \times 100 \times 2.0}{2 \times 0.10} = 4\pi \times 10^{-4} \text{ T}$

Since the currents flow in the opposite sense, the magnetic fields produced by the two coils at their common centre are in opposite directions.

The net magnetic field is $B = |B_1 - B_2| = 4\pi \times 10^{-4} - 4\pi \times 10^{-4} = 0$.