Answer
The concentration of hydroxide ion $(OH^-)$ is equal to $0.020 \space M$, and the pH of that solution is 12.30
Work Step by Step
Since $Ba(OH)_2$ is a strong base with 2 hydroxide ions in it, each mol of this base will produce two moles of $OH^-$:
$$[OH^-] = 2(0.010 \space M) = 0.020 \space M$$
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 0.020 } = 5.0 \times 10^{-13} \space M$$
$$pH = -log[H_3O^+] = -log( 5.0 \times 10^{-13} ) = 12.30 $$
