What is the radius of it's path?

[IHave]

Homework Statement

A 5.6 MeV (kinetic energy) proton enters a 0.18 T field, in a plane perpendicular to the field.
What is the radius of its path?

Homework Equations

kinetic energy: KE = 1/2 mv^2
velocity: v = sqrt(2KE / m)
radius: r = mV/qB

The Attempt at a Solution

KE = 5.6 MeV = 8.9722x10^-13 J
m(p) = 1.67x10^-27 kg
so v = 3.3x10^7 m/s

r = (m(p))(v) / (q(e))(B)

where m(p) is mass of proton
v is velocity as determined by manipulating the KE equation
q(e) is the charge of the proton which is the same as that of an electron (except positive)
B is the given magnetic field

the homework system I'm on, keeps telling me that my answer is wrong. if you see what I'm doing wrong or where I've made a mistake, please let me know.

 

[IHave]

If anyone was in the process of helping me on this one, thank you; but I've figured it out.

I was probably making some sort of calculating error.

r = 1.9 m