A coil having 9 turns carrying current produces magnetic field $\mathrm{B}_1$ at the centre. Now that coil is rewounded into 3 turns carrying same current. Then magnetic field at the centre $B_2$ is

The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 N I}{2R}$, where $N$ is the number of turns, $I$ is the current, and $R$ is the radius.

For the initial coil with $9$ turns, $B_1 = \frac{9\mu_0 I}{2R_1}$.

Rewinding the same length of wire into $3$ turns requires $9R_1 = 3R_2$, so $R_2 = 3R_1$.

The new magnetic field becomes $B_2 = \frac{3\mu_0 I}{2(3R_1)} = \frac{\mu_0 I}{2R_1}$.

Substituting $B_1 = \frac{9\mu_0 I}{2R_1}$ yields $\frac{\mu_0 I}{2R_1} = \frac{B_1}{9}$, hence $B_2 = \frac{B_1}{9}$.

The answer is $\boxed{\text{A}}$.