Step 1: Recall the intensity formula.
In the double slit experiment, the brightness at a point depends on the phase difference $\phi$ between the two waves. The rule is $I = I_0\cos^2\left(\dfrac{\phi}{2}\right)$, where $I_0$ is the brightest possible value.
Step 2: Connect path difference to phase difference.
A path difference $\Delta x$ creates a phase difference $\phi = \dfrac{2\pi}{\lambda}\,\Delta x$. We will use this twice.
Step 3: Handle the first case.
When path difference is one full wavelength $\lambda$, the phase is $\phi = \dfrac{2\pi}{\lambda}\cdot\lambda = 2\pi$. Then $\cos^2(\pi) = 1$, so the intensity is the full $I_0$. We are told this value is $I$, so $I_0 = I$.
Step 4: Handle the second case.
Now the path difference is $\dfrac{\lambda}{4}$. The phase becomes $\phi = \dfrac{2\pi}{\lambda}\cdot\dfrac{\lambda}{4} = \dfrac{\pi}{2}$.
Step 5: Put it in the formula.
The new intensity is $I' = I_0\cos^2\left(\dfrac{\pi}{4}\right)$. Since $\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$, we get $\cos^2\dfrac{\pi}{4} = \dfrac{1}{2}$.
Step 6: Get the final value.
So $I' = I_0\cdot\dfrac{1}{2} = \dfrac{I}{2}$. The brightness at that point falls to half. \[ \boxed{I/2} \]