Answer
$1.86\times10^{-3}\;\rm T$
Work Step by Step
First of all, we need to convert its initial energy to Joules and then find its velocity when its energy is converted to kinetic energy.
According to the conservation of energy,
$$E_i=E_f$$
$$2200\;{\rm eV}=KE$$
Thus,
$$KE=\frac{1}{2}mv^2=2200\cdot 1.6\times 10^{-19}\;\rm J$$
Solving for $v$;
$$v =\sqrt{\dfrac{2\cdot 2200\cdot 1.6\times 10^{-19}}{m}}$$
$$v =\sqrt{\dfrac{2\cdot 2200\cdot 1.6\times 10^{-19}}{9.11\times 10^{-31}}}=\color{blue}{\bf 27.8\times10^6}\;\rm m/s$$
We know that the electron was accelerated horizontally, then entered a magnetic field that exerts a force on it in the $y$-direction.
This means that its velocity in the $x$-direction remains constant.
We also know that the magnetic field forces the electron to move on a circular path (for a small distance which is an arc) which means that the net force exerted on the electron is given by
$$\sum F=F_B=ma_r=\dfrac{mv^2}{r}$$
Thus,
$$F_B=\dfrac{mv^2}{r}$$
whereas $r$ is the radius of the circular path.
We know that the magnetic force is given by $F_B=qvB=evB$ whereas $q=e$ is the charge of the charged particle which is here the electron.
Hence,
$$evB=\dfrac{mv^2}{r}$$
$$B=\dfrac{mv}{er}\tag 1$$
Now we need to find $r$.
$$\tan \theta=\dfrac{11}{22}$$
Thus,
$$\theta=\tan^{-1}\left[\dfrac{11}{22}\right]=\color{blue}{\bf 26.6^\circ}$$
We know that the final velocity vector $v$ at the end of the magnetic field area must be perpendicular to the radius of curvature of the electron path; see the second figure below.
From the geometry of the second figure below, we can see that
$$\sin\theta=\dfrac{3.8}{r}$$
Thus,
$$r =\dfrac{3.8}{ \sin\theta}=\dfrac{3.8}{ \sin26.6^\circ }=\color{blue}{\bf 8.49}\;\rm cm$$
Plugging all of the known into (1);
$$B=\dfrac{mv}{er}=\dfrac{9.11\times10^{-31}\cdot 27.8\times10^6}{1.6\times10^{-19}\cdot 8.49\times10^{-2}}$$
$$B=\color{red}{\bf 1.86\times10^{-3}}\;\rm T$$