Answer
$2.01\times10^{26}atoms/s$
Work Step by Step
We can write the time t that elapses between successive collisions of one atom with one wall of the container is,
$$t=\frac{2L}{v_{rms}}-(1)$$
Since there are three identical pairs of walls, on average 1/3 of the $N_{A}$(1 mol) atoms in the container collide with a given wall in the time t. So,
Collision rate = $\frac{N_{A}}{3t}-(2)$
(1)=>(2),
Collision rate = $\frac{N_{A}}{3(\frac{2L}{v_{rms}})}=\frac{N_{A}v_{rms}}{6L}-(3)$
We know that, $KE_{Avg}=\frac{1}{2}mv_{rms}^{2}=\frac{3}{2}kT=>v_{rms}=\sqrt {\frac{3kT}{m}}$
(4)=>(3),
Collision rate = $\frac{N_{A}}{6L}\sqrt {\frac{3kT}{m}}$ ; Let's plug known values into this equation.
Collision rate = $\frac{6.022\times10^{23}atoms}{6(0.3\space m)}\sqrt {\frac{3(1.38\times10^{-23}J/K)(293\space K)}{3.35\times10^{-26}kg}}\approx2.01\times10^{26}atoms/s$
