Answer
$\omega=24.2rad/s$
Work Step by Step
When the object is attached to the spring, the spring is compressed by $x_0=-0.062m$ and the object's speed $v_0=0$. Finally, when the object is launched at $v_f=1.5m/s$, the spring is back to its unstrained position, so $x_f=0$
According to the conservation of energy, $$KE_0+PE_{elastic, 0}=KE_f+PE_{elastic, f}$$ $$0+\frac{1}{2}kx_0^2=\frac{1}{2}mv_f^2+0$$ $$kx_0^2=mv_f^2$$ $$\frac{k}{m}=\frac{v_f^2}{x_0^2}$$ $$\omega=\sqrt{\frac{k}{m}}=\frac{v_f}{|x_0|}=24.2rad/s$$
