Answer
$\frac{q}{m} = 9.58\times 10^7~C/kg$
This particle is a proton.
Work Step by Step
If the particle passes through without deflection, then $v = \frac{E}{B}$
We can find the charge-to-mass ratio:
$F_B = qvB = \frac{mv^2}{r}$
$\frac{q}{m} = \frac{v}{B~r}$
$\frac{q}{m} = \frac{E}{B^2~r}$
$\frac{q}{m} = \frac{187,500~V/m}{(0.1250~T)^2~(0.12525~m)}$
$\frac{q}{m} = 9.58\times 10^7~C/kg$
We could guess that the charge of the particle is $1.6\times 10^{-19}~C$
We can find the mass:
$\frac{q}{m} = 9.58\times 10^7~C/kg$
$m = \frac{q}{9.58\times 10^7~C/kg}$
$m = \frac{1.6\times 10^{-19}~C}{9.58\times 10^7~C/kg}$
$m = 1.67\times 10^{-27}~kg$
This particle is a proton.
