Answer
$I = 79~mW/m^2$
Work Step by Step
Let $I_1 = 25~mW/m^2$
Let $I_2 = 15~mW/m^2$
We can find the ratio of $\frac{I_1}{I_2}$:
$\frac{I_1}{I_2} = \frac{25~mW/m^2}{15~mW/m^2} = \frac{5}{3}$
The intensity of a wave is proportional to the square of the amplitude, so $\frac{A_1}{A_2} = \sqrt{\frac{5}{3}}$
When two waves interfere constructively, the amplitude of the resulting wave is the sum of the two amplitudes. Therefore, the amplitude of the resulting wave is this case is: $~A_1+A_2 = \sqrt{\frac{5}{3}}~A_2+A_2$
The intensity $I_2$ is proportional to $A_2^2$, while the intensity $I$ of the resulting wave is proportional to $[(1+\sqrt{\frac{5}{3}})~A_2]^2$.
We can find the intensity of the resulting wave:
$I = (1+\sqrt{\frac{5}{3}})^2~I_2$
$I = (1+\sqrt{\frac{5}{3}})^2~(15~mW/m^2)$
$I = 79~mW/m^2$
