The electric field at the center of a circular ring with a volume charge density given by ρ(τ, φ, 2) = ροδ(z)6(-)cos(φ) can be determined by using the principle of superposition and the symmetry of the problem. Let's break down the solution into several steps.

Step 1: Understanding the problem
We are given a circular ring with radius R, centered at the origin. The volume charge density is given by ρ(τ, φ, 2) = ροδ(z)6(-)cos(φ), where ρο is a constant, δ(z) is the Dirac delta function, and φ is the azimuthal angle.

Step 2: Determining the electric field contribution from an infinitesimal charge element
To find the electric field at the center of the ring, we need to consider the contribution from each infinitesimal charge element on the ring. The electric field due to an infinitesimal charge element will be given by Coulomb's law:

dE = k * dq / r²

where dE is the electric field contribution from the infinitesimal charge element, k is the Coulomb's constant, dq is the charge element, and r is the distance between the charge element and the center of the ring.

Step 3: Expressing the infinitesimal charge element in terms of the given charge density
The charge element dq can be expressed in terms of the charge density ρ(τ, φ, 2) and the volume element dτ:

dq = ρ(τ, φ, 2) * dτ

Step 4: Determining the electric field contribution from the entire ring
To find the electric field at the center of the ring, we need to integrate the electric field contribution from each infinitesimal charge element around the entire ring. Since the charge density is only dependent on φ, we can express the volume element dτ as R * dφ * dθ * dr, where R is the radius of the ring, dφ is the infinitesimal angle, dθ is the infinitesimal change in the z-coordinate, and dr is the infinitesimal change in the radial distance.

Step 5: Evaluating the integral
The integral becomes:

E = ∫(0 to 2π) ∫(0 to h) ∫(0 to R) k * ρ(τ, φ, 2) * R * dφ * dθ * dr / r²

where h is the height of the ring (which is negligible in this case since the charge density is given as ροδ(z)).

Step 6: Simplifying the integral and evaluating the electric field
Since the charge density is given as ροδ(z)6(-)cos(φ), the integral simplifies to:

E = k * ρο * R * ∫(0 to 2π) ∫(0 to h) ∫(0 to R) δ(z)6(-)cos(φ) * dφ * dθ * dr / r²

The Dirac delta function δ(z) simplifies the integral over z, and we are left with:

E = k * ρο * R * ∫(0 to 2π) ∫(0 to R) 6(-)cos(φ) * dφ * dr / r²