In triangle $ABC$, $D$ is the midpoint of $AB$; $E$ is the midpoint of $DB$; and $F$ is the midpoint of $BC$. If the area of $\triangle ABC$ is $96$, then the area of $\triangle AEF$ is

$\textbf{(A) }16\qquad\textbf{(B) }24\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad \textbf{(E) }48$
1. **Identify the midpoints and their implications:**
   - \( D \) is the midpoint of \( AB \), so \( AD = DB \).
   - \( E \) is the midpoint of \( DB \), so \( DE = EB \).
   - \( F \) is the midpoint of \( BC \), so \( BF = FC \).

2. **Calculate the area of sub-triangles:**
   - Since \( F \) is the midpoint of \( BC \), the area of \( \triangle ABF \) is half of the area of \( \triangle ABC \). Therefore, the area of \( \triangle ABF \) is:
     \[
     \text{Area of } \triangle ABF = \frac{1}{2} \times 96 = 48
     \]

3. **Further divide the triangle \( \triangle ABF \):**
   - Since \( D \) is the midpoint of \( AB \), the area of \( \triangle ADF \) is half of the area of \( \triangle ABF \). Therefore, the area of \( \triangle ADF \) is:
     \[
     \text{Area of } \triangle ADF = \frac{1}{2} \times 48 = 24
     \]

4. **Calculate the area of \( \triangle BFD \):**
   - Since \( E \) is the midpoint of \( DB \), the area of \( \triangle BFD \) is half of the area of \( \triangle BDF \). Therefore, the area of \( \triangle BFD \) is:
     \[
     \text{Area of } \triangle BFD = \frac{1}{2} \times 24 = 12
     \]

5. **Calculate the area of \( \triangle AEF \):**
   - The area of \( \triangle AEF \) is the area of \( \triangle ABF \) minus the area of \( \triangle BFD \):
     \[
     \text{Area of } \triangle AEF = \text{Area of } \triangle ABF - \text{Area of } \triangle BFD = 48 - 12 = 36
     \]

Conclusion:
\[
\boxed{36}
\]