Let's convert the kinetic energy of the electron from keV to Joule:
[tex]K=45 \cdot 10^3 eV \cdot 1.6\cdot 10^{-19} J/eV =7.2 \cdot 10^{-18}J[/tex]
And from the kinetic energy, we find the velocity of the electron:
[tex]K= \frac{1}{2}mv^2 [/tex]
where [tex]m=9.1 \cdot 10^{-31} kg[/tex] is the electron mass. Re-arranging the formula, we find v:
[tex]v= \sqrt{ \frac{2K}{m} } =3.98 \cdot 10^6 m/s[/tex]
The magnetic force acting on the electron is:
F=qvB
where
[tex]q=1.6 \cdot 10^{-19} C[/tex] is the electron charge
[tex]v=3.98 \cdot 10^6 m/s[/tex] is the electron velocity
B=0.325 T is the magnetic field.
For Newton's second law, we also have that F=ma, so we can rewrite the equation as
[tex]ma=qvB[/tex]
and so from this, we can find the acceleration (which is a centripetal acceleration, because the electron is moving by circular motion):
[tex]a= \frac{qvB}{m}=2.27 \cdot 10^{17} m/s^2 [/tex]
