Rather 

$$\ce{2 NaOH(aq) + CO2(g) -> Na2CO3 (aq) + H2O}$$

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:

$$\ce{CO3^{2-}(aq) + CO2(aq) + H2O <=>> 2 HCO3^- (aq) }$$