Given data

$m=10^{-3}\,\text{kg}, \qquad q=1.0\,\text{C}$ $\vec E(t)=E_0\sin(\omega t)\,\hat i$ with $E_0=1.0\,\text{N C}^{-1}, \qquad \omega=10^3\,\text{rad s}^{-1}$

Force on the particle

Since only electric force acts, $\vec F=q\vec E=qE_0\sin(\omega t)\,\hat i$

Therefore acceleration is $a(t)=\frac{F}{m}=\frac{qE_0}{m}\sin(\omega t)$

Find velocity as a function of time

We use $\frac{dv}{dt}=\frac{qE_0}{m}\sin(\omega t)$

Integrating from $0$ to $t$, and using initial condition $v(0)=0$:

$v(t)=\int_0^t \frac{qE_0}{m}\sin(\omega t')\,dt'$

$v(t)=\frac{qE_0}{m}\left[\frac{1-\cos(\omega t)}{\omega}\right]$

So, $v(t)=\frac{qE_0}{m\omega}\bigl(1-\cos(\omega t)\bigr)$

Maximum value of velocity

Since $-1\le \cos(\omega t)\le 1,$ the quantity $1-\cos(\omega t)$ varies from $0$ to $2$.

Hence maximum speed is $v_{\max}=\frac{qE_0}{m\omega}\times 2$

Substitute values

$v_{\max}=\frac{2(1)(1)}{(10^{-3})(10^3)}$

Since $(10^{-3})(10^3)=1,$ we get $v_{\max}=2\,\text{m s}^{-1}$

Final answer

$\boxed{2}$