It is given that $x$ varies directly as $y$ and inversely as the square of $z$, and that $x=10$ when $y=4$ and $z=14$.  Then, when $y=16$ and $z=7$, $x$ equals:
$ \textbf{(A)}\ 180\qquad\textbf{(B)}\ 160\qquad\textbf{(C)}\ 154\qquad\textbf{(D)}\ 140\qquad\textbf{(E)}\ 120 $
1. Given that \( x \) varies directly as \( y \) and inversely as the square of \( z \), we can write the relationship as:
   \[
   x = k \frac{y}{z^2}
   \]
   where \( k \) is a constant of proportionality.

2. We are given that \( x = 10 \) when \( y = 4 \) and \( z = 14 \). Substituting these values into the equation, we get:
   \[
   10 = k \frac{4}{14^2}
   \]
   Simplifying the denominator:
   \[
   10 = k \frac{4}{196}
   \]
   Solving for \( k \):
   \[
   10 = k \frac{1}{49}
   \]
   \[
   k = 10 \times 49
   \]
   \[
   k = 490
   \]

3. Now, we need to find \( x \) when \( y = 16 \) and \( z = 7 \). Using the equation \( x = k \frac{y}{z^2} \) with \( k = 490 \):
   \[
   x = 490 \frac{16}{7^2}
   \]
   Simplifying the denominator:
   \[
   x = 490 \frac{16}{49}
   \]
   Simplifying the fraction:
   \[
   x = 490 \times \frac{16}{49}
   \]
   \[
   x = 490 \times \frac{16}{49}
   \]
   \[
   x = 16 \times 10
   \]
   \[
   x = 160
   \]

The final answer is \(\boxed{160}\).