A particle executes linear simple harmonic motion with an amplitude...

At position $x$ from the mean position in S.H.M.

velocity, $v = \omega\sqrt{{A}^{2} - {x}^{2}}$

acceleration, $a = x{\omega}^{2}$

here $\omega$ is the angular frequency and $A$ is amplitude.

Given in the problem, $v = a$

$⇒$ $\omega\sqrt{{A}^{2} - {x}^{2}} = x{\omega}^{2}$

$⇒$ $\sqrt{{\left(3\right)}^{2} - {\left(2\right)}^{2}} = 2\left(\dfrac{2\pi}{T}\right)$

$\sqrt{5} = \dfrac{4\pi}{T}$

Time period, $T = \dfrac{4\pi}{\sqrt{5}}$