Assuming that the total pressure $p_\text{total}=2068.44\ \mathrm{kPa}$ is equal to the sum of the partial pressures of the individual gases in accordance with Dalton’s law, the partial pressure of air $p_\text{air}$ can be calculated as follows:
$$\begin{align}
p_\text{total}&=p_\text{air}+p_{\ce{CO2}}\\[6pt]
p_\text{air}&=p_\text{total}-p_{\ce{CO2}}\\[6pt]
&=2068.44\ \mathrm{kPa}-344.74\ \mathrm{kPa}\\[6pt]
&=1723.70\ \mathrm{kPa}
\end{align}$$
Since molar mass $M$ is defined as
$$M=\frac mn$$
where $m$ is mass and $n$ is amount of substance, the ideal gas law
$$p\cdot V = n\cdot R\cdot T$$
where $V$ is volume, $R$ is the molar gas constant, and $T$ is temperature, can be rewritten as
$$\begin{align}
p\cdot V&=\frac{m\cdot R\cdot T}M\\[6pt]
\frac V{R\cdot T}&=\frac m{M\cdot p}
\end{align}$$
This equation applies to air
$$\frac V{R\cdot T}=\frac {m_\text{air}}{M_\text{air}\cdot p_\text{air}}$$
where $m_\text{air}=0.454\ \mathrm{kg}$ is the given mass of air and $M_\text{air}=28.97\ \mathrm{g\ mol^{-1}}$ is the average molar mass of air, as well as to $\ce{CO2}$
$$\frac V{R\cdot T}=\frac {m_{\ce{CO2}}}{M_{\ce{CO2}}\cdot p_{\ce{CO2}}}$$
where $m_{\ce{CO2}}$ is the unknown mass of $\ce{CO2}$ and $M_{\ce{CO2}}=44.01\ \mathrm{g\ mol^{-1}}$ is the molar mass of $\ce{CO2}$.
Therefore,
$$\begin{align}
\frac {m_{\ce{CO2}}}{M_{\ce{CO2}}\cdot p_{\ce{CO2}}}&=\frac {m_\text{air}}{M_\text{air}\cdot p_\text{air}}\\[6pt]
m_{\ce{CO2}}&=\frac {M_{\ce{CO2}}\cdot p_{\ce{CO2}}\cdot m_\text{air}}{M_\text{air}\cdot p_\text{air}}\\[6pt]
&=\frac {44.01\ \mathrm{g\ mol^{-1}}\times344.74\ \mathrm{kPa}\times0.454\ \mathrm{kg}}{28.97\ \mathrm{g\ mol^{-1}}\times1723.70\ \mathrm{kPa}}\\[6pt]
&=0.138\ \mathrm{kg}
\end{align}$$
Note that the given value for the volume $V=85\ \mathrm l$ is not required for the calculation of $m_\text{air}$. Thus, the uncertainty of the volume $V=85\ \mathrm l$ (with only two significant digits) does not affect the uncertainty of the result.
V_total = V_air + V_co2should be,V_total = V_air = V_co2? $\endgroup$