What is the radius of the atoms?
I haven't dived into your math, but I think there are 3 things that are confusing:


intermediate calculations, even though you weren't asked to do any;  
missing units and inconsistent notations (e.g. base 10 is not always shown);  
ambiguous phrase "atoms occupy a volume of $2\times 10^{-4}$."


I always find it cleaner and more productive to derive the formula first, and only then plug in the numbers minding the proper units. As for the volume issue, I tend to believe that $2\times 10^{-4}$ means the fraction of the volume occupied by gas atoms and it is not the absolute value expressed in cubic centimeters. In fact, in this case you get the correct answer and you don't need to use gas volume ($\pu{230 cm3}$) at all.

Check this out. Radius of an atom $r_\mathrm{a}$ is indeed can be found from the volume of an atom $V_\mathrm{a}$

$$r_\mathrm{a} = \sqrt[3]{\frac{3V_\mathrm{a}}{4π}}.$$

At the same time

$$V_\mathrm{a} = \frac{V_\mathrm{atot}}{N}$$

where $V_\mathrm{atot}$ – volume of all gas atoms depicted with spheres; $N$ – number of the atoms that can be found via Avogadro's number $N_\mathrm{A}$ and amount $n$:

$$N = nN_\mathrm{A}.$$

Finally, you are allowed to use ideal gas law to determine the amount $n$:

$$n=\frac{pV}{RT}$$

where $p$ – pressure; $V$ – volume of the gas (macro scale); $R$ – the molar gas constant; $T$ – temperature. Now, lets's put all the things together assuming that  $V_\mathrm{atot}/V = 2\cdot 10^{-4}$:

$$
\begin{align}
r_\mathrm{a} &amp;= \sqrt[3]{\frac{3V_\mathrm{atot}RT}{4πpVN_\mathrm{A}}}\\
  &amp;= \sqrt[3]{\frac{3\cdot2\cdot10^{-4}\cdot\pu{8.314 J mol-1 K-1}\cdot\pu{300 K}}{4\cdot 3.14\cdot\pu{101325 Pa}\cdot\pu{6.022e23 mol-1}}}\\
  &amp;= \pu{1.25e-10 m}
\end{align}
$$

Now that's the value that actually makes sense:)