The concentration of #"HCO"_3^(-)# is #"0.151 mmol/L"#.
The first important aspect about the problem is the fact that you have another reaction that takes places prior to the one given
#CO_(2(g)) rightleftharpoons CO_(2(aq))#
Carbon dioxide gas is dissolved in water, which means that you can determine its concentration by using Henry's law
#P_("gas") = k_H * C_("gas")#, where
#P_("gas")# - the partial pressure of the gas;
#k_H# - Henry's constant - it has a specific value for every gas and it's temperature-dependent;
#C_("gas")# - the concentration of the dissolved gas.
Since your reaction takes place at #"37"^@"C"#, you must calculate the value of #k_H# using a form of the van't Hoff equation (more here: https://chemengineering.wikispaces.com/Henry%27s+Law). I won't detail the calculations because I don't want the answer to become too long. The value you get for #k_H# at #"37"^@"C"# is #"40.0 L" * "atm/mol"#.
This means that the concentration of the dissolved carbon dioxide will be (don't forget to convert mbar to atm)
#[CO_2] = P_(CO_2)/k_H = "0.005181 atm"/("40.0 L" * "atm/mol") = "0.000130 mol/L"#
Side note: at #"25"^@"C"#, the value of #k_H# is #"29.41 L atm/mol"# for carbon dioxide; notice that an increase in temperature reduced the concentration of dissolved carbon dioxide - this is consistent with the idea that gas solubility drops with an increase in temperature.
You now know that you have
#CO_(2(aq)) + H_2O_((l)) rightleftharpoons H_((aq))^(+) + HCO_(3(aq))^(-)#
The expression for the equilibrium constant is
#K = [[H^(+)] * [HCO_3^(-)])/([CO_2]) => [HCO_3^(-)] = (K * [CO_2])/([H^(+)])#
#[HCO_3^(-)] = (0.0000547 * 0.000130)/(0.0000471) = "0.000151 mol/L"#
#[HCO_3^(-)] = "0.151 mmol/L"#
Notice that I've converted all the given concentrations into #"mol/L"# for the calculations.
