Let the amplitude from the slit of width $$d$$ be proportional to $$E$$ and from the slit of width $$xd$$ be proportional to $$xE$$ (with $$x>1$$, as is evident from the given options).

For two coherent waves with amplitudes $$E_1$$ and $$E_2$$, the resultant intensity when added in phase (maximum) and out of phase (minimum) is given by

$$ I_{\max} \propto (E_1+E_2)^2 \quad \text{and} \quad I_{\min} \propto (E_2-E_1)^2. $$

Here, setting

$$E_1 = E$$ (from the narrower slit of width $$d$$) and

$$E_2 = xE$$ (from the wider slit of width $$xd$$),

we have

$$ I_{\max} \propto (E+xE)^2 = E^2 (1+x)^2, $$

and since $$x > 1$$ the subtraction in the destructive case gives

$$ I_{\min} \propto (xE-E)^2 = E^2 (x-1)^2. $$

The ratio of maximum to minimum intensity is therefore

$$ \frac{I_{\max}}{I_{\min}} = \frac{(1+x)^2}{(x-1)^2}. $$

We are given that

$$ \frac{(1+x)^2}{(x-1)^2} = \frac{9}{4}. $$

Taking the square root of both sides (noting that all quantities are positive) leads to

$$ \frac{1+x}{x-1} = \frac{3}{2}. $$

To solve for $$x$$, cross-multiply:

$$ 2(1+x) = 3(x-1). $$

Expanding both sides:

$$ 2 + 2x = 3x - 3. $$

Rearrange the equation to isolate $$x$$:

$$ 2 + 2x + 3 = 3x \quad \Longrightarrow \quad 5 + 2x = 3x, $$

which implies

$$ x = 5. $$

Thus, the value of $$x$$ is

$$ \boxed{5}. $$