Calculate the concentration of iodate ions $\left[\text{IO}_3^-\right]$ in the equilibrium mixture:

The concentration in the prepared solution is given by:

$ \left[\text{IO}_3^-\right]_{\text{eqm}} = \frac{6}{300} = 0.02 \, \text{M} $

Write the expression for the solubility product constant ($ K_{\text{sp}} $) for barium iodate:

$ K_{\text{sp}} = [\text{Ba}^{2+}][\text{IO}_3^-]^2 $

Rearrange to solve for the concentration of barium ions $[\text{Ba}^{2+}]$:

$ [\text{Ba}^{2+}] = \frac{1.58 \times 10^{-9}}{(0.02)^2} = 3.95 \times 10^{-6} \, \text{M} $

The solubility of $\text{Ba}(\text{IO}_3)_2$, which is the same as the concentration of barium ions in this context, is:

$ X = 3.95 $

Thus, the solubility constant $ X $ is $ 3.95 $, and the solubility of barium iodate is $ 3.95 \times 10^{-6} \, \text{mol dm}^{-3} $.