Answer
There is $3.68\times10^{51}~kg$ of "ordinary" matter in the observable universe.
Work Step by Step
First let's calculate the volume of the observable universe:
$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (13.0\times10^{25})^3 = 9.20\times10^{78}~m^3$
The density $\rho = 1\times10^{-26}~kg/m^3$
$mass = V\times\rho = (9.20\times10^{78}~m^3)(1\times10^{-26}~kg/m^3) = 9.20\times10^{52}~kg$
The "ordinary" mass is only 4% of this total mass, so the "ordinary" mass is:
$0.04\times (9.20\times10^{52}~kg) = 3.68\times10^{51} ~kg$
There is $3.68\times10^{51}~kg$ of ordinary matter in the observable universe.
