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22.99 g (1 mol) of Na
12.01          (1 mol) of H
48.0 (1 mol) of C
48.0 (3 mole $\times$ 16.00 gram per mole) of O
The mass of one mole of $NaHCO _3$ is $ 22.99 g + 1.01 g + 12.01 g + 48.00 g = 84.01 g$
And the mass percentages of the elements are
mass % $Na = \dfrac{22.99 g}{84.01 g} \times 100 = 27.36$%
mass % $H = \dfrac{1.01 g}{84.01 g} \times 100 = 1.20$%
mass % $C = \dfrac{12.01 g}{84.01 g} \times 100 = 14.30$%
mass % $O = \dfrac{48.00 g}{84.01 g} \times 100 = 57.14$%