Answer
See the detailed answer below.
Work Step by Step
a) We need to find the mean free path of one electron through an identical gas.
In general, we know that the mean free path of a particle is given by
$$\lambda=\dfrac{1}{4\sqrt{2}\pi (N/V)r^2}\tag 1$$
And it is also given by
$$\lambda=\dfrac{L}{N_{coll}}\tag 2$$
where $L$ is the length of the trajectory and $N_{coll}$ is the number of collisions.
But the author told us that the electron can be thought of as a point particle with zero radius, which means that $r=0$.
So we can not plug that into this formula (1) it will give us an infinite mean-free path of the electron.
We shall follow the same strategy followed by your textbook in section 18.1, but the only difference is the radius between the two particles is there $2r$ while here is only $r$.
The electron here collides with the ideal gas molecules, so then it will collide with any gas molecule that is within $r$ of its path.
Let's assume that the number density of the gas is $N/V$, then the number of collisions of the electron along a path of length $L$ is equal to the number of the gas molecules, and hence it is given by
$$N_{coll}=\dfrac{N}{V}(AL)=\dfrac{N}{V}\pi(r')^2 L$$
where $r'$ is the distance between the center of the electron and the center of the molecule. And since the electron has zero dimensions, $r'=r$
$$N_{coll}= \dfrac{N}{V}\pi r^2 L$$
And a more detailed calculation with all the molecules moving introduces an extra factor of $\sqrt{2}$ since all gas molecules are moving around.
$$N_{coll}= \dfrac{N}{V}\sqrt{2}\;\pi r^2 L$$
Hence,
$$L=\dfrac{N_{coll}}{\sqrt{2}\;\pi r^2 (N/V)}$$
Plugging into (2);
$$\lambda=\dfrac{\dfrac{N_{coll}}{\sqrt{2}\;\pi r^2 (N/V)}}{N_{coll}} $$
Thus,
$$\boxed{\lambda=\dfrac{1}{\sqrt{2}\;\pi r^2 (N/V)}}$$
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b)
We know that the pressure is given by the ideal gas law,
$$PV=Nk_BT$$
Hence,
$$P=k_BT\left[\dfrac{N}{V}\right]\tag 3$$
We are given the mean free path of the electron, so we just need to find $N/V$ so we can then find the pressure inside the tube.
Solving the boxed formula above for $N/V$;
$$\dfrac{N}{V} =\dfrac{1}{\sqrt{2}\;\pi r^2\lambda } $$
Plugging into (3);
$$P=k_BT\left[\dfrac{1}{\sqrt{2}\;\pi r^2\lambda } \right] $$
Plugging the known; and assuming that the gas in the tube is a diatomic gas
$$P=(1.38\times 10^{-23})(20+273)\left[\dfrac{1}{\sqrt{2}\;\pi (1\times 10^{-10})^2(50\times 10^3)} \right] $$
$$P=\color{red}{\bf 1.82\times 10^{-6}}\;\rm Pa=\color{red}{\bf1.8 \times 10^{-11}}\;\rm atm$$
