Answer
$B = 5.0\times 10^{-3}~T$ and this magnetic field is directed out of the page.
Work Step by Step
We can find the magnitude of the electric field between the electrodes:
$E = \frac{\Delta V}{d} = \frac{200~V}{8.0\times 10^{-3}~m} = (2.5\times 10^4)~V/m$
Note that this electric field points from the upper electrode to the lower electrode.
We can find the magnitude of the force exerted on each electron by the electric field:
$F_E = E~q$
$F_E = (2.5\times 10^4~V/m)(1.6\times 10^{-19}~C)$
$F_E = 4.0\times 10^{-15}~N$
For the electrons to pass through undeflected, the force on each electron from the magnetic field must have the same magnitude as $F_E$. We can find the required strength of the magnetic field $B$:
$F_B = q~v~B$
$B = \frac{F_B}{q~v}$
$B = \frac{4.0\times 10^{-15}~N}{(1.6\times 10^{-19}~C)(5.0\times 10^6~m/s)}$
$B = 5.0\times 10^{-3}~T$
Since $F_E$ is directed upward, $F_B$ must be directed downward. By the right hand rule, $B$ must be directed out of the page.
$B = 5.0\times 10^{-3}~T$ and this magnetic field is directed out of the page.
