Calculating volume of gases in different conditions
I haven't found any flaws in your math either, and I can only join those who recommended to include units in your calculations. By googling around I found this problem appeared in 2012 Australian Science Olympiad Exam, and $\pu{13.1 mL}$ was one of the options there. Maybe your source of the answer key is incorrect?

Now to the fun part. I'm pretty sure you were expected to use ideal gas law, but come on, it's Venus and no way under the pressure of $\pu{92 atm}$ any gas would behave even remotely as the ideal one. In fact, $\ce{CO2}$ on Venus with its critical temperature of $\pu{304 K}$ and critical pressure of $\pu{7.39 MPa}$ is going to be in supercritical state!

You have to use Van der Waals' equation:

$$
pV_\mathrm{m} = \frac{RT}{V_\mathrm{m} - b} - \frac{a}{V_\mathrm{m}^2},
$$

where $p$ – pressure; $V_\mathrm{m}$ – molar volume; T – temperature; R – molar gas constant; $a$ and $b$ – characteristic parameters related to the critical temperature $T_\mathrm{cr}$ and pressure $p_\mathrm{cr}$ (reduced form):

$$
a = \frac{27R^2T_\mathrm{cr}^2}{64p_\mathrm{cr}};
\qquad
b = \frac{RT_\mathrm{cr}}{8p_\mathrm{cr}}
$$

Using the above-mentioned values coefficients $a$ and $b$ can be found:

$$
a = \frac{27\cdot (\pu{8.314 J mol-1 K-1})^2\cdot (\pu{304 K})^2}{64\cdot\pu{7.39e6 Pa}} = \pu{3.65e-1 J2 mol-2 Pa-1}
$$

$$
b = \frac{\pu{8.314 J mol-1 K-1}\cdot \pu{304 K}}{8\cdot \pu{7.39e6 Pa}} = \pu{4.28e-5 J mol-1 Pa-1}
$$

Now you need to plug both values in the first equation and solve quadratic equation for $V_\mathrm{m}$. Wolfram Alpha claims $V_\mathrm{m} = \pu{2.56e-2 m3 mol-1}$, so

$$
V(\ce{CO2}) = \pu{2.56e-2 m3 mol-1}\cdot\pu{2e-2 mol} = \pu{5.12e-1 L},
$$

which, of course, deviates quite a lot from the value found using ideal gas law.