Number of grams of CO2 released from a carbonated water
This shows part of the problem. 

Henry's law as given is 

$K_{AW}^{px} = $1.65*10^3$ = \frac{p_{\ce{CO2(gas)atm}}}{[\ce{CO2(aq)}]}$

where:
$K_{AW}^{px}$ atm = aqueous-phase mixing ratio of $\ce{CO2}$ and the units are $atm$

$[\ce{CO2{(aq)}}]$ = the concentration of $\ce{CO2{(aq)}}$ in $mole/L$

$p_{\ce{CO2(gas)atm}}]$ = the partial pressure of $\ce{CO2{(gas)}}$ in $atm$ 

So rearranging...

$[\ce{CO2(aq)}] =\frac{p_{\ce{CO2(gas)atm}}}{K_{AW}}$

The moles of $\ce{CO2}$, $m_{\ce{CO2}}$, is given by concentration, $[\ce{CO2}]$, times the volume, $v$ in $L$.

$m_{\ce{CO2(aq)}} = [\ce{CO2(aq)}]v =  \frac{p_{\ce{CO2(gas)atm}}}{K_{AW}}v$