Calculate Kc with dissociation degree
You don't need gas laws, temperature or $K_p.$
You overcomplicated the problem: $K_c$ can be found by following its definition with an RICE table:
$$
\begin{array}{lccccc}
&amp;\text{R}\ce{&amp;2 HI(g) &amp;&lt;=&gt; &amp;H2(g) &amp;+ &amp;I2(g)} \\
&amp;\text{I} &amp; 2c_0 &amp;&amp; 0 &amp;&amp; 0 \\
&amp;\text{C} &amp; -2\alpha c_0 &amp;&amp; \alpha c_0 &amp;&amp; \alpha c_0 \\
&amp;\text{E} &amp; 2(1 - \alpha)c_0 &amp;&amp; \alpha c_0 &amp;&amp; \alpha c_0 \\
\end{array}
$$
For the sake of consistency, I denoted the initial concentration with $c_0$ and dissociation degree as $\alpha.$
By definition, the equilibrium constant on a concentration basis $K_c$ can be found as follows:
$$K_c = \frac{[\ce{H2(g)}][\ce{I2(g)}]}{[\ce{HI(g)}]^2} = \frac{(\alpha c_0)^2}{(2(1 - \alpha)c_0)^2} = \frac{\alpha^2}{4(1 - \alpha)^2} = \frac{0.2198^2}{4(1 - 0.2198)^2} \approx 0.0198$$