At what distance along the z-axis is the electric field strength a maximum?

The axial field is the integration of the field from each element of charge around the ring. Because of symmetry, the field is only in the direction of the axis. The field from an element ds in the ring is 

dE = (qs*ds)cos(T)/(4*pi*e0)*(x^2 + R^2) 

where x is the distance along the axis from the plane of the ring, R is the radius of the ring, qs is the linear charge density, T is the angle of the field from the x-axis. 

However, cos(T) = x/sqrt(x^2 + R^2) 

so the equation becomes 

dE = (qs*ds)*[x/sqrt(x^2 + R^2)]/(4*pi*e0)*(x^2 + R^2) 

dE =[qs*ds/(4*pi*e0)]*x/(x^2 + R^2)^1.5 

Integrating around the ring you get 

E = (2*pi*R/4*pi*e0)*x/(x^2 + R^2)^1.5 

E = (R/2*e0)*x*(x^2 + R^2)^-1.5 

we differentiate wrt x, the term R/2*e0 is a constant K, and the derivative is 

dE/dx = K*{(x^2 + R^2)^-1.5 +x*[(-1.5)*(x^2 + R^2)^-2.5]*2x} 

dE/dx = K*{(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5} 

to find the maxima set this = 0, giving 

(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0 

mult both side by (x^2 + R^2)^2.5 to get 

(x^2 + R^2) - 3*x^2 = 0 

-2*x^2 + R^2 = 0 

-2*x^2 = -R^2 

x = (+/-)R/sqrt(2) 

2.73 K

Explanation:

The Cosmic Microwave Background (CMB) radiation are a type of electromagnetic radiations that are released after the  explosion of the big bang. It propagates in all direction in space and is moving further with the expansion of the universe. They are not visible but they are helpful in estimating the age of the universe.

These radiations have a wavelength of about 970 μm (approximately 1 mm).

The temperature of the corresponding blackbody of this radiation is approximately 2.73 K(-270.42°C) and its extremely cold.