Answer
$1.83\;\rm Hz$
Work Step by Step
We know that after the block hits the spring it then oscillates with an amplitude of 10 cm from its new equilibrium point.
The new equilibrium point is below the initial equilibrium point since the weight of the ball compresses the spring down.
See the figure below.
This means that the spring will stretch up to a maximum amplitude of 10 cm from its new (final) equilibrium point after the collision and compresses down to a maximum distance of 10 cm below its new equilibrium point. And this process goes on forever (in theory).
So, we can see that the block undergoes simple harmonic motion.
Now we need to find the spring constant, we can use the new equilibrium and applying Newton's second law.
The net force exerted on the block at the equilibrium point is zero, so
$$ kx_1=mg $$
Hence,
$$k=\dfrac{mg}{x_1}\tag 1$$
We know that the angular frequency is given by
$$\omega=2\pi f= \sqrt{\dfrac{k}{m}}$$
Plugging from (1) and solving for $f$;
$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{\color{red}{\bf\not} mg}{\color{red}{\bf\not} mx_1}}$$
$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{ g}{ x_1}}\tag 2$$
Now we need to find $x_1$ so we can find the frequency magnitude.
We can use the conservation of energy law. Initially, the block was 3 cm above the spring and finally it is at a distance of $x_1+A$, as seen in the figure below, from the initial equilibrium point of the spring.
So,
$$E_1i=E_f$$
$$K_i+U_{gi}+U_{si}=K_f+U_{gf}+U_{sf}$$
Initially the spring was on its normal length, and finally the ball stops.
$$K_i+U_{gi}+0=0+0+U_{sf}$$
we chose the point at which the ball stops to be the zero level at which $y=0$, so $y_i= x_1+A$
$$\frac{1}{2}mv_i^2+mg( x_1+A)=\frac{1}{2}k(x_1+A)^2 $$
where $x_1+A$ is the compressed distance by the spring when the ball stopped.
$$ mv_i^2+2mg( x_1+A)= k(x_1+A)^2\tag{$\times 2$}$$
Plugging from (1);
$$ \color{red}{\bf\not} mv_i^2+2\color{red}{\bf\not} mg( x_1+A)= \dfrac{\color{red}{\bf\not} mg}{x_1}(x_1+A)^2 $$
$$ v_i^2+2 g( x_1+A)= \dfrac{ g}{x_1}(x_1+A)^2 $$
Plugging the known and solving for $x_1$;
$$ v_i^2+2 (9.8)( x_1+0.10)= \dfrac{ 9.8}{x_1}(x_1+0.10)^2 $$
;;;;;
$$ v_i^2+19.6x_1+1.96= \dfrac{ 9.8}{x_1}(x_1^2+0.2x_1+0.01 )$$
$$ v_i^2+19.6x_1+1.96= 9.8 x_1 +1.96+\dfrac{0.098}{x_1} $$
$$ v_i^2+9.8 x_1= \dfrac{0.098}{x_1} $$
$$ v_i^2x_1+9.8 x_1^2 -0.098 =0\tag 3 $$
Now we need to find $v_i$ which is the velocity of the block just before it hits the spring which we can find by the kinematic formula of
$$v_i^2=v_0^2+2g\Delta y$$
$$v_i^2=0^2+2(9.8)(0.03)=0.588$$
Plugging into (3);
$$ 9.8 x_1^2+ 0.588x_1 -0.098 =0 $$
Thus,
$$x=-0.134\;{\rm m}\;\;\;\;{\rm or},\;\;\;x=\bf 0.0744 \;\rm m$$
We need to cancel the negative root since it is greater than $A$ in magnitude.
Plug the positive root into (2);
$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{ g}{ x_1}} =\dfrac{1}{2\pi}\sqrt{\dfrac{ 9.8}{ 0.0744 }} $$
$$f=\color{red}{\bf 1.83}\;\rm Hz$$
