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}

Using the diameter of the muzzle $d$ and the initial distance between the bullet and the end of the muzzle $l$ we can calculate the volume of the gas to be given by: \begin{equation} V\left(l\right)=\frac{\pi d^2\left(L-l\right)}{4} \end{equation} where $L$ is the total length of the muzzle. We therefore find from the ideal gas law that: \begin{equation} \begin{aligned} P\left(l\right)&=\frac{nRT}{V\left(l\right)}\\ &=\frac{4nRT}{\pi d^2\left(L-l\right)} \end{aligned} \end{equation} We equivalently have in terms of the initial pressure $P\left(l_0\right)=\frac{4nRT}{\pi d^2\left(L-l_0\right)}$ that: \begin{equation} P\left(l\right)=\frac{P\left(l_0\right)\left(L-l_0\right)}{L-l} \end{equation} Assuming the bullet fills the cross-sectional area of the muzzle we therefore find that the force the gas exerts on the bullet is given by: \begin{equation} \begin{aligned} F\left(l\right)&=P\left(l\right)A\\ &=\frac{\pi d^2P\left(l_0\right)\left(L-l_0\right)}{4\left(L-l\right)} \end{aligned} \end{equation} Assuming the bullet initially begins at rest we find that: \begin{equation} \begin{aligned} W&=-\int_{l_0}^0F\left(l\right)\, \textrm{d}l\\ \implies\frac{1}{2}mv^2&=\frac{\pi d^2P\left(l_0\right)\left(L-l_0\right)}{4}\int_0^{l_0}\frac{\textrm{d}l}{L-l}\\ \implies v&=d\sqrt{\frac{\pi P\left(l_0\right)\left(L-l_0\right)\ln\frac{L}{L-l_0}}{2m}}\\ \end{aligned} \end{equation} 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&\approx 0.006\sqrt{\frac{\pi\left(1.824\times 10^8\right)\left(0.8-0.01\right)\ln\frac{0.8}{0.8-0.01}}{2\left(0.002\right)}}\\ &\approx 226.4\textrm{ m s}^{-1} \end{aligned} \end{equation}