How do I calculate the molarity of the Na2CO3 solution?
The first part makes more sense writing moles explicitly. The first step is determining how many moles of sodium carbonate are in the original $1.00\ \mathrm  g$ sample:

$$\frac{1.00\ \mathrm{g}}{286\ \mathrm{g/mol}} = 0.003447\ \mathrm{mol}$$

The second step is determining the concentration of the first $20.0\ \mathrm{mL}$ solution, i.e. what concentration results from dissolving the $0.00345\ \mathrm{mol}$ of sodium carbonate calculated above.
$$\frac{0.003447\ \mathrm{mol}}{0.0200\ \mathrm{L}} = 0.174825\ \mathrm{M}$$

The third part is simply a dilution – what concentration results when we dilute the first $20.0\ \mathrm{mL}$ solution to $250\ \mathrm{mL}$:
$$c_2 = \frac{c_1V_1}{V_2}$$
$$c_2 = \frac{(0.174825\ \mathrm{M})(0.020\ \mathrm{L})}{(0.250\ \mathrm{L})} = 0.0140\ \mathrm{M}$$