Decompose the velocity into a component perpendicular to $\vec{B}$ (which gives circular motion of radius $R$) and a component parallel to $\vec{B}$ (which advances the helix).

From $R = \dfrac{m v_\perp}{eB}$:

$$v_\perp = \dfrac{e}{m} B R = (1.76 \times 10^{11})(2 \times 10^{-3})(0.20) = 7.04 \times 10^{7} \text{ m/s}.$$

The pitch equals the parallel distance traveled in one cyclotron period $T = \dfrac{2\pi m}{eB}$:

$$v_\parallel = \dfrac{h}{T} = \dfrac{h \cdot (e/m) \cdot B}{2\pi} = \dfrac{(0.05)(1.76 \times 10^{11})(2 \times 10^{-3})}{2\pi} \approx 2.80 \times 10^{6} \text{ m/s}.$$

Total speed: $v = \sqrt{v_\perp^2 + v_\parallel^2} \approx 7.05 \times 10^{7}$ m/s.