Answer
$T = 0.421~s$
Work Step by Step
We can use conservation of momentum to find the speed of the block just after the bullet strikes it. Let $M$ be the total mass of the block and the bullet. Let $m_b$ be the mass of the bullet.
$Mv_2 = m_bv_1$
$v_2 = \frac{m_bv_1}{M}$
$v_2 = \frac{(0.00800~kg)(280~m/s)}{1.00~kg}$
$v_2 = 2.24~m/s$
We can find the force constant of the spring.
$\frac{1}{2}kA^2 = \frac{1}{2}Mv_{max}^2$
$k = \frac{Mv_{max}^2}{A^2}$
$k = \frac{(1.00~kg)(2.24~m/s)^2}{(0.150~m)^2}$
$k = 223~N/m$
We can find the period.
$T = 2\pi~\sqrt{\frac{M}{k}}$
$T = 2\pi~\sqrt{\frac{1.00~kg}{223~N/m}}$
$T = 0.421~s$