1. Key observation about the geometry

   In an equilateral triangle, the triangle formed by joining the midpoints of the three sides is also an equilateral triangle.

   So, triangle $DEF$ is similar to triangle $ABC$ with linear scale factor $k=\frac{1}{2}.$

   Hence:

   • area scales as $k^2=\frac14$
   • mass scales as $\frac14$ (uniform sheet)
   • distances from the common centroid scale as $\frac12$

2. Centroid of the medial triangle

   Since $D,E,F$ are the midpoints of the sides of $ABC$, triangle $DEF$ is the medial triangle. Its centroid is the same as the centroid $G$ of triangle $ABC$.

   Therefore, the axis through $G$ perpendicular to the plane is also the centroidal perpendicular axis for triangle $DEF$.

3. Scaling of moment of inertia

   For similar laminae about corresponding centroidal axes, $I \propto M L^2.$

   Here,

   • $M' = \frac14 M$
   • $L' = \frac12 L$

   Thus the moment of inertia of triangle $DEF$ about the same centroidal perpendicular axis is

   =\frac14\cdot\frac14 I_0 =\frac{1}{16}I_0.$$

4. Moment of inertia of the remaining part

   The remaining figure is obtained by removing triangle $DEF$ from triangle $ABC$.

   So, $I=I_0-I_{DEF}=I_0-\frac{1}{16}I_0=\frac{15}{16}I_0.$

5. Option check

   $I=\frac{15}{16}I_0$ matches Option B.

Final Answer: $\boxed{I=\frac{15}{16}I_0}$