Finding initial amount of ammonia using equilibrium constant and amount of nitrogen gas
Since the volume is given and it's a constant, the initial amount of ammonia $n_0(\ce{NH3})$ can be found from its initial concentration $c_0(\ce{NH3}):$

$$n_0(\ce{NH3}) = c_0(\ce{NH3})\times V\tag{1}$$

To find $c_0(\ce{NH3}),$ an ICE table might indeed come in handy; however, yours needs corrections.
First, I suggest to rewrite it according to the process occurring in the system, namely dissociation of ammonia:

$$
\begin{array}{lccc}
\ce{&amp;2 NH3(g) &amp;&lt;=&gt; &amp;N2(g) &amp;+ &amp;3 H2(g)} \\
\text{I} &amp; c_0 &amp;&amp; 0 &amp;&amp; 0 \\
\text{C} &amp; -2x &amp;&amp; +x &amp;&amp; +3x \\
\text{E} &amp; c_0 - 2x &amp;&amp; x &amp;&amp; 3x \\
\end{array}
$$

Second, note that the provided equilibrium constant $K_c$ is given for the synthesis of ammonia, hence in our case a reciprocal value is relevant:

$$
\begin{align}
\frac{1}{K_c} &amp;= \frac{[\ce{N2}][\ce{H2}]^3}{[\ce{NH3}]^2}\tag{2.1}\\
\frac{1}{K_c} &amp;= \frac{x (3x)^3}{(c_o - 2x)^2}\tag{2.2}\\
\frac{1}{K_c} &amp;= \frac{27x^4}{(c_o - 2x)^2}\tag{2.3}\\
\frac{1}{\sqrt{K_c}} &amp;= \frac{3\sqrt{3}x^2}{c_o - 2x}\tag{2.4}
\end{align}
$$

$$c_0 = 3\sqrt{3K_c}x^2 + 2x\tag{3}$$

At the equilibrium there is $\pu{0.200 mol}$ of nitrogen in the $\pu{1 L}$ vessel, so $x = \pu{0.200 mol L-1},$ and the initial amount of ammonia can be found as follows:

$$
\begin{align}
n_0(\ce{NH3}) &amp;= (3\sqrt{3K_c}x^2 + 2x) × V\\
  &amp;= (3\sqrt{3\times 4.20}\times (\pu{0.200 mol L-1})^2 + 2\times \pu{0.200 mol L-1})\times \pu{1 L} \\
  &amp;= \pu{0.826 mol}\tag{4}
\end{align}
$$