Rather
$$\ce{2 NaOH(aq) + CO2(g) -> Na2CO3 (aq) + H2O}$$
and analogically on the solid $\ce{NaOH}$.
Bicarbonate in aqueous solutions cannot survive the excess of hydroxide, forming carbonate.
$$\ce{HCO3^{-}(aq) + OH- (aq)<=>> CO3^{2-}(aq) + H2O}$$
The same is further enforced by dehydration effect of solid hydroxide.
$$\ce{NaHCO3(s) + NaOH(s) -> Na2CO3(s) + H2O}$$
Only in excess of $\ce{CO2}$ Is formed bicarbonate:
$$\ce{CO3^{2-}(aq) + CO2(aq) + H2O <=>> 2 HCO3^- (aq) }$$