Textbook problem with molar mass of carbon dioxide
The problem is that, when the flask is empty, it weighs $85.431\ \mathrm g$. But this weight is the weight of the empty flask plus the air included. So the first thing to do is to calculate the weight of the air, and the corresponding number of moles of air in the flask before adding $\ce{CO2}$. The volume of the air is the same as the volume of water when it is filled with water, namely $122\ \mathrm{mL}$.
The amount of air is $$n = pV/RT = \frac{122\times10^{-6}\ \mathrm{m^3}\times101325\ \mathrm{Pa}}{8.316\times300\ \mathrm{J/mol}} = 4.956\times10^{-3}\ \ \mathrm{mol}$$ The mass of this air is : $$m = 4.956\times10^{-3}\ \mathrm{mol}\times28.8\ \mathrm{g/mol} = 0.1427\ \mathrm g$$ So the mass $m_0$ of the empty flask without air is : $$m_0 = 85.431\ \mathrm g - 0.1427\ \mathrm g = 85.2883\ \mathrm g$$
A s consequence, the mass of the $\ce{CO2}$ in the flask is : $$m(\ce{CO2}) = 85.510\ \mathrm g - 85.2883\ \mathrm g = 0.2227\ \mathrm g$$
The molar mass of this $\ce{CO2}$ is: $$M = \frac{m}{n} = \frac{0.2227\ \mathrm g}{4.956\times10^{-3}\ \mathrm{mol}}= 44.8\ \mathrm{g/mol}$$
This is nearly the expected value for the molar mass of $\ce{CO2}$. If you take into account the fact that the volume $122\ \mathrm{mL}$ is known with a precision $\pm1\ \mathrm{mL}$, or $\pm1\ \%$, the final result will be defined with an uncertainty of $\pm1\ \mathrm{g/mol}$. It is: $$M(\ce{CO2}) = 44.8\ \mathrm{g/mol} \pm 1\ \mathrm{g/mol}$$