Answer
$6.55\times10^{25}\;\rm collision/s$
Work Step by Step
We know that the average force due to molecule collision is given by
$$F_{avg}=2\dfrac{ N_{\rm coll}}{\Delta t} mv_x$$
Hence the rate of collisions is given by
$$\dfrac{ N_{\rm coll}}{\Delta t} =\dfrac{F_{avg}}{2mv_x}$$
Recalling that $F_{avg}=PA$ where $P$ is the pressure and $A$ is the cross-sectinoal area.
$$\dfrac{ N_{\rm coll}}{\Delta t} =\dfrac{PA}{2mv_x}$$
we are given the pressure $P$, the cross-sectional area of the wall $A$, and we can find the mass of the nitrogen molecule.
where the mass of the nitrogen molecule is given by
$$m_{N_2}=2M_N (1.66\times 10^{-27}) $$
where $M_N$ is the atomic mass of the nitrogen atom.
So,
$$\dfrac{ N_{\rm coll}}{\Delta t} =\dfrac{PA}{4M_N (1.66\times 10^{-27}) v_x}\tag 1$$
So we have to find $v_x$:
Noting that
$$v_x=\sqrt{(v_x^2)_{avg}}$$
and since the rms speed is for the 3 dimensions, then $(v_x^2)_{avg}=v_{\rm rms}^2/3$
$$v_x=\sqrt{\dfrac{v_{\rm rms}^2}{3}}=\dfrac{v_{\rm rms}}{\sqrt3}$$
Plugging into (1);
$$\dfrac{ N_{\rm coll}}{\Delta t} =\dfrac{\sqrt3PA}{4M_N (1.66\times 10^{-27}) v_{\rm rms}} $$
Plugging the known;
$$\dfrac{ N_{\rm coll}}{\Delta t} =\dfrac{\sqrt3(2\times 1.013\times 10^5)(0.1\times 0.1)}{4(14) (1.66\times 10^{-27}) (576)} $$
$$\dfrac{ N_{\rm coll}}{\Delta t} =\color{red}{\bf 6.55\times10^{25}}\;\rm collision/s$$
