$$ M_{max} = \frac{wL^2}{8} = \frac{2 \times (20)^2}{8} = 100 \text{ kip-ft} $$ (Convert to kip-in: 100 * 12 = 1,200 kip-in)
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You need to select a W-shape steel beam (A992 steel, Fy=50 ksi) to span 20 feet, supporting a uniform load of 2 kips/ft (including self-weight). Deflection is limited to L/360. $$ M_{max} = \frac{wL^2}{8} = \frac{2 \times (20)^2}{8}
$$ Z_{req} = \frac{M_{max}}{\phi F_y} = \frac{1,200 \text{ kip-in}}{0.9 \times 50 \text{ ksi}} = 26.7 \text{ in}^3 $$ Deflection is limited to L/360