You should know that centripedal force is F = mv^2/r (mass times centripedal acceleration a = v^2/r). For your problem, m is the mass of the proton, v is the proton's velocity (the unknown you need to solve for), and r is 10^6 m plus the Earth's radius.
But you need the left-side of the equation. How does magnetic force relate to a magnetic field strength? Use the Right-Hand Rule: make your index finger, middle finger, and thumb all perpendicular to each other. Your index finger is the direction of the magnetic field B. Your thumb is the direction of the proton's velocity v. And your middle finger is the direction of the magnetic force F. For your situation, where the charge is moving perpendicular to B, the equation for force is F = qvB. Thus
qvB = mv^2/r
Solve for v...
v = mv^2 / qBr
v / v^2 = m / qBr
1/v = m / qBr
v = qBr / m
q = charge of a proton = 1.6 x 10^-19 C
B = 4 x 10^-8 T
r = r_earth + 10^6 m = 7.37 x 10^6 m
m = mass of a proton = 1.67 x 10^-27 kg
Plugging everything in...
v = (1.6 x 10^-19 C)(4 x 10^-8 T)(7.37 x 10^6 m) / (1.67 x 10^-27 kg)
v = 2.82 x 10^7 m/s
That's almost 9% of the speed of light!