Assuming the bullet fills the cross-sectional area of the muzzle the gas exerts a force of:
\begin{equation}
\begin{aligned}
F&=PA\\
&=\frac{\pi d^2P}{4}
\end{aligned}
\end{equation}
where $d$ is the diameter of the muzzle. Assuming the bullet initially begins at rest we find from the work-energy theorem (where $T$ is the kinetic energy) that:
\begin{equation}
\begin{aligned}
W&=T_{\textrm{final}}-T_{\textrm{initial}}\\
\implies Fl&=\frac{1}{2}mv^2\\
\implies v&=\sqrt{\frac{2Fl}{m}}\\
&=\frac{d}{2}\sqrt{\frac{2\pi Pl}{m}}
\end{aligned}
\end{equation}
where $l$ is the distance the bullet travels in the muzzle and $v$ is the bullet's velocity at the end of the muzzle. Substituting our given values (as $1800\textrm{ atm}\approx 1.824\times 10^8\textrm{ Pa}$) we therefore find that:
\begin{equation}
\begin{aligned}
v&=\frac{d}{2}\sqrt{\frac{2\pi Pl}{m}}\\
&\approx\frac{0.006}{2}\sqrt{\frac{2\pi\left(1.824\times 10^8\right)\left(0.01\right)}{0.002}}\textrm{ m s}^{-1}\\
&\approx 227.1\textrm{ m s}^{-1}
\end{aligned}
\end{equation}