Number of collisions of particles in a box (or a sphere)
If you want a number of collisions in a given time $\Delta t$, then simply from a units perspective, the other answer doesn't make sense, as it produces a number with dimensions of length. You want something that is dimensionless.

Often, physical chemistry textbooks will provide a derivation for collisions per unit volume per unit time. The proofs are a little bit involved and I'd rather refer you to a textbook instead of typing it all out. However, the bottom line is that the collision frequency is given by

$$Z_{AB} = \pi(r_A + r_B)^2\left[\langle v_A \rangle^2 + \langle v_B \rangle^2 \right]^{1/2} \left(\frac{N_A}{V}\frac{N_B}{V}\right)$$


$Z_{AB}$ is the number of collisions between A and B per unit time per unit volume
$r_A$ and $r_B$ are the radii of A and B
$\langle v_A \rangle^2$ and $\langle v_B \rangle^2$ are the mean square velocities of A and B
$N_A$ and $N_B$ are the number of particles of A and B
$V$ is the total volume


For a derivation of this I suggest reading Levine's Physical Chemistry, 6th ed, section 14.7.

For your specific case you've assumed that $r_B = 0$, $v_A = 0$, $v_B = s$. Therefore this simplifies to

$$Z_{AB} = \pi r_A^2 \cdot s \cdot \left(\frac{N_A}{V}\frac{N_B}{V}\right)$$

Obviously, the radius of A, $r_A$ is related to its volume $V_A$ by $V_A = 4\pi r_A^3 /3$. To obtain the number of collisions in a given time you just need to multiply by the total volume and total time.

$$f =  \pi r_A^2 \cdot s\Delta t \cdot \left(\frac{N_A N_B}{V}\right)$$

As you can see, the form of this equation is not all too different from the one in the other answer. The key difference is that the distance travelled $s\Delta t$ must be multiplied by a cross-sectional collision area, equal to $\pi r_A^2$, and not by a volume. The textbooks will explain this well, I will try to elaborate on this later if I have time.