We know that for a normal population, \begin{align} \frac{(n-1)S^{2}}{\sigma^{2}} \sim \chi_{n-1}^{2} \end{align}

In the given problem, $n = 15$ and $\sigma^{2} = 9$. Hence, \begin{align} P(S^{2} > 12) &= P(\frac{14 S^{2}}{9} > \frac{14}{9} 12)\newline &= P(\chi_{14}^{2} > \frac{56}{3})\newline &= 0.178 \quad\text{from standard tables or chi-square calculators} \end{align}