Given $ B= 3.5 \;T$
$q= 1.6 \times 10^{-19}$
$m= 1.67 \times 10^{-27}$
$v= \large\frac{B_q}{2 \pi m} =\frac{3.5 \times 1.6 \times 10^{-19}}{2 \times \Large\frac{22}{7} \times 1.67 \times 10^{27}}$
$\qquad= 5.35 \times 10^7 \;H^z$
Answer : $ 5.35 \times 10^7 \;Hz$