Let the widths of the two slits be $w_1$ and $w_2$. Given $w_1 : w_2 = 1 : 25$. The intensity of light coming from a slit is directly proportional to its width ($I \propto w$). Therefore, the ratio of intensities is $\frac{I_1}{I_2} = \frac{1}{25}$. Also, intensity is directly proportional to the square of the amplitude ($I \propto a^2$). So, $\frac{a_1^2}{a_2^2} = \frac{1}{25} \implies \frac{a_1}{a_2} = \frac{1}{5}$. Let $a_1 = x$ and $a_2 = 5x$. The maximum intensity in the interference pattern is $I_{max} \propto (a_1 + a_2)^2 = (x + 5x)^2 = (6x)^2 = 36x^2$. The minimum intensity is $I_{min} \propto (a_1 - a_2)^2 = (x - 5x)^2 = (-4x)^2 = 16x^2$. The ratio of maximum to minimum intensity is: $\frac{I_{max}}{I_{min}} = \frac{36x^2}{16x^2} = \frac{36}{16} = \frac{9}{4}$.