The motion of the particle is described by the three equations, 

$x=4\sin{2t}$,

$y=4\cos{2t}$

and $z=6t$

So the displacement of the particle can be written as,

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$or,\ \vec{r}= 4\sin{2t}\hat{i}+4\cos{2t}\hat{j}+6t\hat{k}$

where, $\hat{i}$, $\hat{j}$ and $\hat{k}$ are the unit vectors in three dimensional Cartesian co-ordinate system along the X-axis, Y-axis and Z-axis respectively.

Again the velocity is the time derivative of the displacement vector, so the velocity is given by,

$\displaystyle{\vec{v}=\frac{d\vec{r}}{dt}}$

$or,\ \displaystyle{\vec{v}=\frac{d}{dt}[4\sin{2t}\hat{i}+4\cos{2t}\hat{j}+6t\hat{k}]}$

$or,\ \vec{v}=4(\cos{2t})\times{2}\hat{i}+4(-\sin{2t})\times{2}\hat{j}+6\hat{k}$

$or,\ \vec{v}=8\cos{2t}\hat{i}-8\sin{2t}\hat{j}+6\hat{k}$

So the magnitude of the velocity is

$v=|\vec{v}|=\sqrt{8^2{\cos}^2{2t}+8^2{\sin}^2{2t}+6^2}$

$or,\ v=\sqrt{8^2({\cos}^2{2t}+{\sin}^2{2t})+6^2}$

$or,\ v=\sqrt{8^2+6^2}$

$or,\ v=\sqrt{100}=10\ m/sec$

Now the acceleration is given by,

$\vec{a}=\frac{d\vec{v}}{dt}$

$or,\ \vec{a}=\frac{d}{dt}[8\cos{2t}\hat{i}-8\sin{2t}\hat{j}+6\hat{k}]$

$or,\ \vec{a}=8(-\sin{2t})\times{2}\hat{i}-8(\cos{2t})\times{2}\hat{j}$

$or,\ \vec{a}=-16\sin{2t}\hat{i}-16\cos{2t}\hat{j}$

So the magnitude of the acceleration is given by,

$a=|\vec{a}|=\sqrt{16^2{\sin}^2{2t}+16^2{\cos}^2{2t}}$

$or,\ a=\sqrt{16^2({\sin}^2{2t}+{\cos}^2{2t})}=\sqrt{16^2}$

$or,\ a=16\ m/{sec}^2$