Answer
$ 8.9\times 10^{24}\;\rm collision/s$
Work Step by Step
Let's assume that the wall we will work on is a vertical wall, so the collisions are in the $x$-direction.
We know that the rate of collision is given by
$$\dfrac{N_{coll}}{\Delta t}=\frac{1}{2}\dfrac{N}{V}Av_x\tag 1$$
Now we need to find $v_x$ in terms of the rms speed since this equation is right if all the molecules are moving in the $x$-direction at the same time while the rms speed is for the 3 directions.
Recalling that
$$v_x=\sqrt{(v_x)^2_{avg} }=\sqrt{\dfrac{v_{\rm rms}^2}{3}}$$
Hence,
$$v_x=\dfrac{v_{\rm rms}}{\sqrt{3}}$$
Plugging into (1);
$$\dfrac{N_{coll}}{\Delta t}=\frac{1}{2}\dfrac{N}{V}\dfrac{v_{\rm rms}}{\sqrt{3}}A$$
$$\dfrac{N_{coll}}{\Delta t}=\frac{A}{2\sqrt{3}}\dfrac{N}{V} v_{\rm rms} \tag 2$$
Recalling that $v_{\rm rms}$ is given by
$$v_{\rm rms}=\sqrt{\dfrac{3k_BT}{m}}$$
where $m$ here is the mass of the nitrogen molecule which is given by $m=2M_N(1.661\times 10^{-27})$
$$v_{\rm rms}=\sqrt{\dfrac{3k_BT}{2M_N(1.661\times 10^{-27})}}$$
Plugging into (2);
$$\dfrac{N_{coll}}{\Delta t}=\frac{A}{2\sqrt{3}}\dfrac{N}{V} \sqrt{\dfrac{3k_BT}{2M_N(1.661\times 10^{-27})}}\tag 3$$
Now we need to find the number of molecules $N$ which is given by
$N=nN_{\rm A}$ and $V$ is given by $V=L^3$. Also $A=L^2$
Plugging into (3);
$$\dfrac{N_{coll}}{\Delta t}=\frac{L^2}{2\sqrt{3}}\dfrac{nN_{\rm A}}{L^3} \sqrt{\dfrac{3k_BT}{2M_N(1.661\times 10^{-27})}} $$
hence,
$$\dfrac{N_{coll}}{\Delta t}=\frac{1}{2\sqrt{3}}\dfrac{nN_{\rm A}}{L} \sqrt{\dfrac{3k_BT}{2M_N(1.661\times 10^{-27})}} $$
Plugging the known;
$$\dfrac{N_{coll}}{\Delta t}=\frac{1}{2\sqrt{3}}\dfrac{ (0.01)(6.02\times 10^{23})}{(0.1)} \sqrt{\dfrac{3 (1.38\times 10^{-23})(20+273)}{2(14)(1.661\times 10^{-27})}} $$
$$\dfrac{N_{coll}}{\Delta t}=\color{red}{\bf 8.9\times 10^{24}}\;\rm collision/s$$