The normal distribution deals with continuous probability, so to accommodate a discrete value we need to calculate the probabilities for 22.5 and 23.5 and then subtract. We need two Z scores:
Z1=(22.5-18)/6=4.5/6=0.75 and Z2=(23.5-18)/6=5.5/6=0.9167.
p(Z1)=0.7734, p(Z2)=0.8203. Therefore p(X=23)=0.8203-0.7734=0.0469 or 4.69% approx.