1. Understand the geometry

   In an equilateral triangle, joining the midpoints of the sides forms the medial triangle. So triangle $DEF$ is also equilateral and is similar to triangle $ABC$ with linear scale factor $k=\frac{1}{2}.$

   Hence:

   • side of $DEF = \frac{1}{2}$ side of $ABC$
   • area of $DEF = \left(\frac12\right)^2 = \frac14$ of area of $ABC$

2. Location of the centroid

   For an equilateral triangle, the centroid is also the centre of symmetry of the medial triangle formed by joining midpoints. Therefore, the centroid of cavity $DEF$ is the same point $O$ as that of the original triangle $ABC$.

   So the axis through $O$ perpendicular to the plane passes through the centroid of both triangles.

3. Use scaling of moment of inertia

   For similar laminae of the same uniform surface density, moment of inertia about corresponding centroidal perpendicular axes scales as $I \propto (\text{mass})(\text{length})^2.$

   Since mass scales with area, mass ratio is $\frac{m_{DEF}}{m_{ABC}} = \frac14.$

   Also, squared length scale is $\left(\frac12\right)^2 = \frac14.$

   Therefore, $\frac{I_{DEF}}{I_{ABC}} = \frac14 \cdot \frac14 = \frac{1}{16}.$

   Given $I_{ABC}=I_0$, we get $I_{DEF}=\frac{I_0}{16}.$

4. Moment of inertia of remaining lamina

   The remaining part is obtained by removing the cavity $DEF$ from the original lamina, so

   = \frac{15 I_0}{16}.$$

5. Check options

   • A: $\frac78 I_0$ ❌
   • B: $\frac{15}{16} I_0$ ✅
   • C: $\frac34 I_0$ ❌
   • D: $\frac{31}{32} I_0$ ❌

Thus the correct answer is $\boxed{\frac{15}{16}I_0}.$