Answer
$0.41\;\rm s$
Work Step by Step
We know that the position of an object that undergoes a simple harmonic motion is given by
$$x_{(t)}=A\cos(\omega t+\phi_0)\tag 1$$
At $t=0$, $x=0$, so
$$0=A\cos(\omega (0)+\phi_0)=A\cos (\phi_0)$$
Thus,
$$\phi_0=\cos^{-1}(0)=\pm \dfrac{\pi }{2}$$
And since the object is traveling to the right, so it must be in the fourth quadrant, so that
$$\phi_0= \dfrac{-\pi }{2}\tag 2$$
Plugging into (1);
$$x_{(t)}=A\cos\left(2\pi f t+ \dfrac{ -\pi }{2}\right) $$
$$x_{(t)}=A\cos\left(\frac{2\pi}{T} t- \dfrac{ \pi }{2}\right) $$
Plugging the known;
$$6=10 \cos\left(\frac{2\pi}{4} t- \dfrac{ \pi }{2}\right) $$
$$ \cos\left(\frac{ \pi}{2} t- \dfrac{ \pi }{2}\right) =0.6$$
Recall that $\cos(\theta-90^\circ)=\sin\theta$, so
$$ \sin\left(\frac{ \pi}{2} t \right) =0.6$$
$$ \frac{\pi}{2} t =\sin^{-1}(0.6)$$
$$ t =\dfrac{2\sin^{-1}(0.6) }{\pi }=\color{red}{\bf 0.41}\;\rm s$$
