Answer
$^{22}_{11}Na + \;^{2}_{1}H\rightarrow\;^{20}_{10}Ne + \alpha$.
Work Step by Step
$$^{22}_{11}Na + \;^{2}_{1}H\rightarrow\;? + \alpha$$
Recall that $ \alpha$ is $^{4}_{2}He$.
Conserve nucleon number: 22 + 2 = A + 4. We see that A = 20.
Conserve charge: 11 + 1 = Z + 2. We see that Z = 10.
$$^{22}_{11}Na + \;^{2}_{1}H\rightarrow\;^{20}_{10}Ne + \alpha$$