Strontium hydroxide dissociates as follows: $$ \mathrm{Sr(OH)}_2 \rightarrow \mathrm{Sr}^{2+} + 2\mathrm{OH}^- $$ Given the concentration of $\mathrm{Sr(OH)}_2$ is $0.0003 \text{ M}$, the concentration of hydroxide ions is: $$ [\mathrm{OH}^-] = 2 \times 0.0003 \text{ M} = 0.0006 \text{ M} $$ Now, calculate the pOH: $$ \begin{aligned} \text{pOH} &= -\log[\mathrm{OH}^-]\\ &= -\log(0.0006)\\ &= 3.2218 \end{aligned} $$ Finally, calculate the pH: $$ \begin{aligned} \text{pH} &= 14 - \text{pOH}\\ &= 14 - 3.2218\\ &= 10.7782 \end{aligned} $$ The pH of the solution is $10.778$.