1. Set up the idea using negative mass

Let the center of the original circular disk be at $O$, taken as the origin.

The circular disk is uniform, so if a square hole is removed, we can treat the removed square as a negative mass located at its own center.

Since the square hole is at distance $d=\frac a2$ from $O$, the center of the square lies at $x=\frac a2$ on, say, the positive $x$-axis.

Then the center of mass of the remaining part will lie on the same axis and be shifted to the opposite side, so its coordinate will be negative.

2. Mass of the full disk

Let surface mass density be $\sigma$.

Radius of disk $=a$, so $M_{\text{disk}}=\sigma \pi a^2.$

3. Mass of the removed square

Side of square hole: $l=\frac a2.$

So area of square is $l^2=\left(\frac a2\right)^2=\frac{a^2}{4}.$

Hence mass removed is $m_{\text{sq}}=\sigma \frac{a^2}{4}.$

4. Center of mass of remaining portion

Using the formula with removed mass treated as negative: $x_{\text{CM}}=\frac{M_{\text{disk}}\cdot 0 - m_{\text{sq}}\cdot d}{M_{\text{disk}}-m_{\text{sq}}}.$

Substitute values: $x_{\text{CM}}=\frac{-\left(\sigma \frac{a^2}{4}\right)\left(\frac a2\right)}{\sigma \pi a^2-\sigma \frac{a^2}{4}}.$

Cancel $\sigma a^2$: $x_{\text{CM}}=\frac{-\frac a8}{\pi-\frac14}.$

Now simplify: $x_{\text{CM}}=-\frac{a}{8\left(\pi-\frac14\right)}=-\frac{a}{8\pi-2}.$

So the distance is of the form $-\frac{a}{X},$ therefore $X=8\pi-2.$

5. Numerical value

Using $\pi\approx 3.1416$, $X=8\pi-2\approx 8(3.1416)-2=25.1328-2=23.1328.$

Nearest integer: $\boxed{23}.$

6. Comparison with stored answer

Derived answer = $23$

Stored correct answer = $23$

They agree.