Twenty cubical blocks are arranged as shown. First, $10$ are arranged in a triangular pattern; then a layer of $6$, arranged in a triangular pattern, is centered on the $10$; then a layer of $3$, arranged in a triangular pattern, is centered on the $6$; and finally one block is centered on top of the third layer. The blocks in the bottom layer are numbered $1$ through $10$ in some order. Each block in layers $2, 3$ and $4$ is assigned the number which is the sum of the numbers assigned to the three blocks on which it rests. Find the smallest possible number which could be assigned to the top block.
[asy]
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$ \textbf{(A)}\ 55 \qquad\textbf{(B)}\ 83 \qquad\textbf{(C)}\ 114 \qquad\textbf{(D)}\ 137 \qquad\textbf{(E)}\ 144 $
To solve this problem, we need to understand the structure of the blocks and how the numbers are assigned to each block in the upper layers based on the blocks below them. Let's break down the problem step by step.

1. **Label the Blocks in the Bottom Layer:**
   We label the blocks in the bottom layer (Layer 1) as follows:
   \[
   \begin{array}{cccc}
   A & B & C & D \\
   E & F & G \\
   H & I \\
   J \\
   \end{array}
   \]
   This gives us 10 blocks in the bottom layer.

2. **Calculate the Numbers for the Second Layer:**
   Each block in the second layer is the sum of the three blocks directly below it. We label the blocks in the second layer (Layer 2) as follows:
   \[
   \begin{array}{ccc}
   K & L & M \\
   N & O \\
   P \\
   \end{array}
   \]
   The numbers assigned to these blocks are:
   \[
   \begin{aligned}
   K &= A + B + E, \\
   L &= B + C + F, \\
   M &= C + D + G, \\
   N &= E + F + H, \\
   O &= F + G + I, \\
   P &= H + I + J.
   \end{aligned}
   \]

3. **Calculate the Numbers for the Third Layer:**
   Each block in the third layer is the sum of the three blocks directly below it. We label the blocks in the third layer (Layer 3) as follows:
   \[
   \begin{array}{cc}
   Q & R \\
   S \\
   \end{array}
   \]
   The numbers assigned to these blocks are:
   \[
   \begin{aligned}
   Q &= K + L + N, \\
   R &= L + M + O, \\
   S &= N + O + P.
   \end{aligned}
   \]

4. **Calculate the Number for the Top Block:**
   The top block (Layer 4) is the sum of the three blocks directly below it. We label the top block as \( T \):
   \[
   T = Q + R + S.
   \]

5. **Express \( T \) in Terms of the Bottom Layer Blocks:**
   We need to express \( T \) in terms of the blocks in the bottom layer. Substituting the expressions for \( Q, R, \) and \( S \):
   \[
   \begin{aligned}
   Q &= (A + B + E) + (B + C + F) + (E + F + H), \\
   R &= (B + C + F) + (C + D + G) + (F + G + I), \\
   S &= (E + F + H) + (F + G + I) + (H + I + J).
   \end{aligned}
   \]
   Simplifying these:
   \[
   \begin{aligned}
   Q &= A + 2B + C + 2E + 2F + H, \\
   R &= B + 2C + D + 2F + 2G + I, \\
   S &= E + 2F + G + 2H + 2I + J.
   \end{aligned}
   \]
   Adding these together to get \( T \):
   \[
   \begin{aligned}
   T &= (A + 2B + C + 2E + 2F + H) + (B + 2C + D + 2F + 2G + I) + (E + 2F + G + 2H + 2I + J) \\
   &= A + 3B + 3C + D + 3E + 6F + 3G + 3H + 3I + J.
   \end{aligned}
   \]

6. **Minimize \( T \):**
   To find the smallest possible value of \( T \), we assign the smallest values to the blocks in the bottom layer. The smallest values are \( 1, 2, 3, \ldots, 10 \). Assigning these values optimally:
   \[
   \begin{aligned}
   A &= 1, \\
   B &= 2, \\
   C &= 3, \\
   D &= 4, \\
   E &= 5, \\
   F &= 6, \\
   G &= 7, \\
   H &= 8, \\
   I &= 9, \\
   J &= 10.
   \end{aligned}
   \]
   Substituting these values into the expression for \( T \):
   \[
   \begin{aligned}
   T &= 1 + 3(2) + 3(3) + 4 + 3(5) + 6(6) + 3(7) + 3(8) + 3(9) + 10 \\
   &= 1 + 6 + 9 + 4 + 15 + 36 + 21 + 24 + 27 + 10 \\
   &= 153.
   \end{aligned}
   \]

However, we need to check if there is a mistake in the calculation or if there is a more optimal way to assign the values. Let's re-evaluate the optimal assignment:

Assigning the values such that the sum is minimized:
\[
\begin{aligned}
A &= 10, \\
B &= 9, \\
C &= 8, \\
D &= 7, \\
E &= 6, \\
F &= 5, \\
G &= 4, \\
H &= 3, \\
I &= 2, \\
J &= 1.
\end{aligned}
\]
Substituting these values into the expression for \( T \):
\[
\begin{aligned}
T &= 10 + 3(9) + 3(8) + 7 + 3(6) + 6(5) + 3(4) + 3(3) + 3(2) + 1 \\
&= 10 + 27 + 24 + 7 + 18 + 30 + 12 + 9 + 6 + 1 \\
&= 144.
\end{aligned}
\]

Thus, the smallest possible number which could be assigned to the top block is \( \boxed{144} \).