Missing questions: "find the speed of the electron".
Solution:
the magnetic force experienced by a charged particle in a magnetic field is given by
[tex]F=qvB \sin \theta[/tex]
where
q is the particle charge
v its velocity
B the magnitude of the magnetic field
[tex]\theta[/tex] the angle between the directions of v and B.
Re-arranging the formula, we find:
[tex]v= \frac{F}{qB \sin \theta} [/tex]
and by substituting the data of the problem (the charge of the electron is [tex]q=1.6 \cdot 10^{-19} C[/tex]), we find the velocity of the electron:
[tex]v= \frac{F}{qB \sin \theta}= \frac{4.90 \cdot 10^{-15}N}{(1.6 \cdot 10^{-19}C)(3.70 \cdot 10^{-3} T)(\sin 60^{\circ})}=9.56 \cdot 10^6 m/s [/tex]
