Answer
${\bf 2.93}\;\rm mT $
Work Step by Step
The electron undergoes a cyclotron motion that has a radius that is given by
$$r=\dfrac{mv}{qB}$$
So the magnetic field strength is given by
$$B=\dfrac{m_ev}{er}\tag 1$$
Now we need to find $v$, so we can use the conservation of energy.
$$E_i=E_f$$
$$K_i+U_{ie}=K_f+U_{fe}$$
$$0+0=\frac{1}{2}m_v^2-eV$$
Hence,
$$v=\sqrt{\dfrac{2eV}{m_e}} $$
Plug into (1),
$$B=\dfrac{m_e }{er}\sqrt{\dfrac{2eV}{m_e}}$$
$$B=\sqrt{\dfrac{2 Vm_e}{r^2e }}\tag 2$$
Now we need to find the radius of curvature of the electron motion.
From the geometry of the figure below, note the right triangle below,
$$\dfrac{r}{\sin90^\circ}=\dfrac{0.02}{\sin\theta}$$
Hence,
$$r=0.02/\sin\theta$$
Plug into (2);
$$B=\sqrt{\dfrac{2 Vm_e}{ 0.02^2 e (\sin\theta)^2}} $$
Plug the known;
$$B=\sqrt{\dfrac{2(10\times 10^3)(9.11\times 10^{-31}) (\sin10^\circ)^2 }{ 0.02^2 (1.6\times 10^{-19}) }} $$
$$B=\color{red}{\bf 2.93}\;\rm mT $$
