1. Analyze the Geometry of the Star Loop

1. The conducting wire loop is a symmetric star with 6 outer vertices and 6 inner vertices. It is composed of 12 identical straight wire segments.
2. Let the center of the star be the origin O. The distance between diametrically opposite outer vertices is given as $4a$. Therefore, the distance from the center O to any outer vertex (let's call it P) is $R_{outer} = OP = \frac{4a}{2} = 2a$.
3. The 12 wire segments connect the outer and inner vertices alternately (e.g., $P_1 \to Q_1 \to P_2 \to \dots$). Due to the symmetry, the 12 triangles formed by connecting the center O to the endpoints of each segment (like $\triangle OP_1Q_1$, $\triangle OQ_1P_2$, etc.) are all congruent.
4. The total angle around the center O is $360^\circ$. Since this angle is divided into 12 equal parts by the vertices, the angle subtended by each segment at the center is $\angle P_1OQ_1 = \frac{360^\circ}{12} = 30^\circ$.

2. Determine the Specific Geometry

1. To solve for the magnetic field, we need more information about the shape, specifically the angles of the triangle $\triangle OP_1Q_1$. We are given $OP_1 = 2a$ and $\angle P_1OQ_1 = 30^\circ$. A reasonable assumption for a symmetric star is that the angle at each outer point is $60^\circ$. Let's denote the outer vertices as $P_1, P_2, \dots, P_6$ and inner vertices as $Q_1, Q_2, \dots, Q_6$. The angle at vertex $P_1$ is $\angle Q_6P_1Q_1$.
2. Let's assume $\angle Q_6P_1Q_1 = 60^\circ$. By symmetry, the line $OP_1$ bisects this angle. Therefore, $\angle OP_1Q_1 = \frac{60^\circ}{2} = 30^\circ$.
3. Now, consider the triangle $\triangle OP_1Q_1$. We have two angles: $\angle P_1OQ_1 = 30^\circ$ and $\angle OP_1Q_1 = 30^\circ$. This means $\triangle OP_1Q_1$ is an isosceles triangle.
4. The third angle is $\angle OQ_1P_1 = 180^\circ - (30^\circ + 30^\circ) = 120^\circ$.

3. Calculate the Magnetic Field from a Single Segment

1. The magnetic field at a point due to a finite straight wire carrying current $I$ is given by: $B = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2)$ where $d$ is the perpendicular distance from the point to the wire, and $\theta_1, \theta_2$ are the angles subtended by the ends of the wire from the foot of the perpendicular.
2. Let's find the perpendicular distance $d$ from the center O to the wire segment $P_1Q_1$. Let H be the foot of the perpendicular from O to the line containing $P_1Q_1$. In $\triangle OP_1H$, we have: $d = OH = OP_1 \sin(\angle OP_1Q_1) = (2a) \sin(30^\circ) = 2a \left(\frac{1}{2}\right) = a$
3. Since $\angle OQ_1P_1 = 120^\circ$ (obtuse), the foot of the perpendicular H lies outside the segment $P_1Q_1$. Therefore, the magnetic field is given by the difference of the contributions: $B_{segment} = \frac{\mu_0 I}{4\pi d} |\sin\theta_{far} - \sin\theta_{near}|$ where the angles are measured from the perpendicular OH to the lines $OP_1$ and $OQ_1$.
4. Let's find these angles. In the right-angled triangle $\triangle OHP_1$:
   • $\angle OP_1H = 30^\circ$ (as established).
   • $\theta_{far} = \angle P_1OH = 90^\circ - 30^\circ = 60^\circ$.
5. To find $\theta_{near} = \angle Q_1OH$, we can use the fact that $\angle P_1OQ_1 = 30^\circ$. Both $P_1$ and $Q_1$ lie on the same side of the perpendicular OH.
   • $\angle Q_1OH = \angle P_1OH - \angle P_1OQ_1 = 60^\circ - 30^\circ = 30^\circ$.
6. Now, substitute the values into the formula for $B_{segment}$: $B_{segment} = \frac{\mu_0 I}{4\pi a} (\sin 60^\circ - \sin 30^\circ)$ $B_{segment} = \frac{\mu_0 I}{4\pi a} \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) = \frac{\mu_0 I}{4\pi a} \left(\frac{\sqrt{3}-1}{2}\right)$

4. Calculate the Total Magnetic Field

1. The star has 12 identical segments. By the right-hand rule, the current in each segment produces a magnetic field at the center O in the same direction (e.g., out of the page).
2. The total magnetic field $B_{total}$ is the sum of the magnitudes of the fields from all 12 segments: $B_{total} = 12 \times B_{segment}$ $B_{total} = 12 \times \frac{\mu_0 I}{4\pi a} \left(\frac{\sqrt{3}-1}{2}\right)$ $B_{total} = \frac{\mu_0 I}{4\pi a} \times 6 (\sqrt{3}-1)$

5. Conclusion

The magnitude of the magnetic field at the center of the loop is $\frac{\mu_0 I}{4\pi a} 6[\sqrt{3}-1]$. This matches option A.