In the given nuclear reaction, the element $X$ is:
$${}_{11}^{22}\text{Na}\to X+e^{+}+\nu$$

The reaction represents positron ($e^{+}$) decay.A positron decay follows the general form:
$${}_Z^{A}P{\to }_{Z-1}^{A}D+e^{+}+\nu$$
Applying this to Sodium (${}_{11}^{22}\text{Na}$):Mass number $A$ remains $22$.Atomic number $Z$ becomes $11-1=10$.The element with $Z=10$ is Neon ($\text{Ne}$).So, $X$ is ${}_{10}^{22}\text{Ne}$.