A particle of mass m and charge q is projected from the origin with velocity v0x in a region where a uniform magnetic field Bz and a uniform electric…

Along $z$ the particle has constant acceleration $a=qE/m$, starting with zero $z$-velocity. In the $xy$-plane the magnetic force alone acts, so the motion is a circle traversed at constant speed $v_0$ with radius $mv_0/(qB)$, independent of $E$; (A) is correct. The time for one turn, $2\pi m/(qB)$, depends only on $m,q,B$, so (B) is wrong. The $z$-displacement is $z=\tfrac12at^2$. In the first period it is $\tfrac12aT_0^2$; in the second it is $\tfrac12a(4T_0^2-T_0^2)=\tfrac32aT_0^2$, three times the first, so (C) is correct (a numerical integration with $E\neq0$ gave a ratio $3.000$). After one turn $v_z=aT_0\neq0$, so the speed is $\sqrt{v_0^2+(aT_0)^2}>v_0$ and (D) is wrong. Correct options: A and C.