The gas will occupy 120 L at STP.
We can use the Combined Gas Laws .
In 1982, chemists redefined STP as 1 bar and 0 °C (273.15 K).
p_1 = 5100 mmHg × (1 atm/760 mmHg) × (1.013 bar/1 atm) = 6.798 bar
T_1 = (29 + 273.15) K = 302.15 K
p_1V_1/T_1 = p_2V_2/T_2
V_2 = V_1 × p_1/p_2 × T_2/T_1
= 20.1 L × (6.798 bar/1 bar) × (273.15 K/302.15 K) = 120 L