The volume of this asteroid is required in cubic centimeters, so we need to convert the given diameter in km to a radius in cm.  Therefore the required radius is:
r = 20,000 * 100 / 2 = 1,000,000 cm
Now assuming we are looking at a spherical NEO, the volume will be:
V = 4 * π * (1000000)$$³ / 3 = 4.188790205 X 10$$¹⁹ cm$$³
This is a big assumption because the asteroid could be elliptical where the diameter we see, is across the objects semiminor axis, in which case the object's actual volume will be greater than calculated by an unknown amount.  However we could just as easily be looking across the asteroid's semimajor axis, in which case the astroid's actual volume will be less than the calculated value by an unknown amount.  For elliptical astroids with small eccentricities, the spherical volume is a valid approximation of the asteroid's actual volume. Unfortunately we have no way to measure its eccentricity unless it is tumbling through space so that we can observe its changing diameter.
A first approximation of the asteroid's mass for an H class asteroid will be:
3.8 (g/cm$$³) * 4.188790205 X 10$$¹⁹ (cm$$³) / 1000 (g/Kg) = 1.591740979 X 10$$¹⁷ Kilograms
Converting this to tonnage:
1.591740979 X 10$$¹⁷ (Kg) * 2.204622476 (lb/Kg) = 3.509187938 X 10$$¹⁷ pounds, and:
3.509187938 X10$$¹⁷(lb) / 2204.6 (lb/metric ton) = 1.591757207 X 10$$¹⁴ metric tons
1.591757207 X 10$$¹⁴ / 1.0 X 10 $$⁶ (metric tons/metric megatons) = 1.59 X 10$$⁸ = 159 Million Metric Megatons