Answer
$F = 7.4 \times 10^{-12}~N$ (directed to the east)
Work Step by Step
We can find the speed of the proton:
$\frac{1}{2}mv^2 = K$
$v^2 = \frac{2K}{m}$
$v = \sqrt{\frac{2K}{m}}$
$v = \sqrt{\frac{(2)(8.0\times 10^{-13}~J)}{1.67\times 10^{-27}~kg}}$
$v = 3.095\times 10^7~m/s$
We can find the magnetic force:
$F = q~v\times B$
$F = (1.6\times 10^{-19}~C)(3.095\times 10^7~m/s)(1.5~T)$
$F = 7.4 \times 10^{-12}~N$
By the right hand rule, $v \times B$ is directed east.
