Let the equilateral triangle have side length $a$ and total mass $M$.

Since the lamina is uniform, mass is proportional to area.

Moment of inertia of the full triangle about centroid $G$.

Given: $I_G = I_0$

Identify the removed part $\triangle ADE$.

Since $D$ and $E$ are midpoints of $AB$ and $AC$, triangle $ADE$ is similar to triangle $ABC$ with linear scale factor $\frac12$.

Hence, $\text{Area}(ADE)=\frac14\text{Area}(ABC)$ so its mass is $m=\frac{M}{4}.$

Moment of inertia of the small triangle about its own centroid.

For similar laminae, moment of inertia scales as $I \propto (\text{mass})(\text{length})^2.$

Here mass scales by $\frac14$ and length by $\frac12$, so $I_{\text{small, centroid}}=\frac14\cdot\left(\frac12\right)^2 I_0=\frac{I_0}{16}.$

Find distance between centroid of $ABC$ and centroid of $ADE$.

Place coordinates: $A(0,0),\quad B(a,0),\quad C\left(\frac a2,\frac{\sqrt3 a}{2}\right).$

Then $G=\left(\frac a2,\frac{\sqrt3 a}{6}\right).$

Midpoints: $D\left(\frac a2,0\right),\quad E\left(\frac a4,\frac{\sqrt3 a}{4}\right).$

Centroid of $\triangle ADE$ is $g=\frac{A+D+E}{3}=\left(\frac a4,\frac{\sqrt3 a}{12}\right).$

Therefore, $\vec{Gg}=\left(\frac a4,\frac{\sqrt3 a}{12}\right),$ and

Use parallel axis theorem for the removed triangle about axis through $G$.

$I_{ADE\text{ about }G}=I_{\text{small, centroid}}+m(Gg)^2.$

So,

Relate $I_0$ to $Ma^2$.

For an equilateral triangular lamina of mass $M$ and side $a$, about centroidal axis perpendicular to its plane: $I_0=\frac{Ma^2}{12}.$

Therefore, $\frac{M}{4}\cdot\frac{a^2}{12}=\frac{Ma^2}{48}=\frac{I_0}{4}.$

Hence, I_{ADE\text{ about }G}=\frac{I_0}{16}+\frac{I_0}{4}=rac{5I_0}{16}.

Moment of inertia of remaining part.

$I_{\text{remaining}}=I_0-\frac{5I_0}{16}=\frac{11I_0}{16}.$

Comparing with $I_{\text{remaining}}=\frac{N I_0}{16},$ we get $N=11.$

Compare with stored answer.

Stored correct answer = $11$, which matches our result.