Step 1: Picture the motion as a drift plus circular motion.
In crossed uniform fields $\vec E = E\hat x$ and $\vec B = B\hat z$, a charged particle's velocity always splits into a steady drift velocity $\vec v_d$ plus a circular rotation at the cyclotron frequency $\omega = qB/m$ around that drift. Here $\vec v_d = \vec E \times \vec B / B^2$ points along $-\hat y$ with size $E/B$.
Step 2: Locate the start on the velocity circle.
The particle starts at rest, $\vec v(0) = 0$. In velocity space this is a point on a circle of radius $E/B$ centred at $\vec v_d = (0,-E/B)$; the origin sits at distance $E/B$ directly above this centre, so the starting velocity is at the top of the circle. As time passes, the velocity vector sweeps around this circle at the steady rate $\omega$.
Step 3: Relate $v_x$ to the swept angle.
Measuring the angle $\phi = \omega t$ swept from the start, $v_x(t) = \frac{E}{B}\sin(\omega t)$: positive for $0 < \omega t < \pi$, negative for $\pi < \omega t < 2\pi$, and it returns to zero once every half circle ($\omega t = \pi, 2\pi, 3\pi,\dots$).
Step 4: Connect this to the return on the $y$-axis.
The $x$-displacement is the running area under $v_x(t)$: $x(t) = \frac{1}{\omega}\int_0^{\omega t} \frac{E}{B}\sin\phi\, d\phi = \frac{E}{B\omega}(1-\cos\omega t)$. This is zero only when $\cos\omega t = 1$, a full turn, $\omega t = 2\pi, 4\pi,\dots$ (the half turn at $\omega t = \pi$ gives $x = 2E/(B\omega) \neq 0$, the farthest point of the loop, not a return). So the particle is back on the $y$-axis only after one complete revolution of the velocity vector, at $t = 2\pi/\omega$.
Step 5: Put in the numbers.
\[ \omega = \frac{qB}{m} = \frac{(3.0\times10^{-4}\text{ C})(3\text{ T})}{9.0\times10^{-5}\text{ kg}} = 10\text{ rad/s} \]
\[ t = \frac{2\pi}{10} = 0.6283\text{ s} \]
Final Answer:
This rounds to
\[ \boxed{t = 0.63\text{ s}} \]