Answer
$\bf proton$,
${\bf 9.581\times 10^7}\;\rm C/kg$
Work Step by Step
We have here two fields that force the electron beam to move in a straight line. The electric field pushes the electron downward while the magnetic field pushes the electron upward or vice versa.
This means that the two forces exerted by the two fields are equal in magnitude.
Thus,
$$F_B=F_E$$
$$qvB\sin90^\circ=qE$$
Thus,
$$v=\dfrac{E}{B} \tag 1$$
Now we have a zero electric field just outside the plates, as we see in the figure below, so the magnetic field deflects the electron into a circular path that has a radius of
$$r=\dfrac{mv}{qB} $$
Hence,
$$\dfrac{q}{m}=\dfrac{v}{rB}$$
Plug from (1),
$$\dfrac{q}{m}=\dfrac{E}{rB^2}$$
Plug the known;
$$\dfrac{q}{m}=\dfrac{(187\;500)}{(\frac{25.05}{2}\times 10^{-2})(0.125)^2}$$
$$\dfrac{q}{m}=\color{red}{\bf 9.581\times 10^7}\;\rm C/kg$$
Let's test, the mass of the electron and the proton since both have the same amount of charge.
$$\dfrac{e}{m_e}=\dfrac{1.6\times 10^{-19}}{9.11\times 10^{-31}}=\bf 1.75631\times 10^{11}\;\rm C/kg$$
So it is not the electron,
$$\dfrac{e}{m_p}=\dfrac{1.6\times 10^{-19}}{1.67\times 10^{-27}}=\bf 9.58084\times 10^7\;\rm C/kg$$
Therefore, the particle is a $\bf proton$.
