A glowing sign holds 0.020 mol of neon at 330 K…

Step 1. One atom. $E = \tfrac32 k_BT = \tfrac32 \times 1.38 \times 10^{-23}\ \text{J/K} \times 330\ \text{K} = 6.83 \times 10^{-21}\ \text{J}$.

Step 2. All the neon. $U = \tfrac32 nRT = \tfrac32 \times 0.020\ \text{mol} \times 8.31\ \text{J/(mol K)} \times 330\ \text{K} = 82.3\ \text{J}$.