(a) The entire cluster subtends an angle of 9 arcminutes.  Since 1 radian 
equals 57.3 degrees and 1 degree = 60 arcminutes, the diameter of the cluster

  d = r * (9 arcmin) / (57.3 deg) = 867 Mpc * (9 / (57.3 * 60)) = 2.3 Mpc

Since the Milky Way is about 30 kpc (or 0.030 Mpc) in diameter, the cluster
Abell 2218 is 2.3/0.3 = 77 times larger in diameter than our galaxy.


(b) If we approximate the shape of Abell 2218 to be spherical, its volume is

  V = (4/3) * pi * (d/2)^3 = 4.19 * (2.3/2)^3 = 6.2 Mpc^3

We also know that the Milky Way galaxy has approximately 10^11 stars, at about
2 * 10^30 kg per star.  So, if each of the 120 galaxies in this cluster's
volume is the mass of the Milky Way, then the VISIBLE mass (excluding dark
matter) of the cluster would total

  M = 120 * 10^11 * (2 * 10^30 kg) = 2.4 * 10^43 kg

So the density of VISIBLE mass of the cluster would be

  D = M/V = (2.4 * 10^43 kg) / (6.2 Mpc^3) = 3.9 * 10^42 kg/Mpc^3

or, in more conventional units, we recall that 1 pc = 3.1 * 10^16 m; so

  1 Mpc^3 = (10^6 * 3.1 * 10^16 m)^3 = 3.0 * 10^67 m^3  and

  D = (3.9 * 10^42) / (3.0 * 10^67) = 1.3 * 10^(-25) kg/m^3

This density is ten trillion trillion times lower than air at sea level,
or a little less than a hundred hydrogen atoms per cubic meter.