The maximum intensity in a Young's double slit experiment is I_0 ....

The position of the point on the screen exactly opposite to one of the slits is at a distance $y = \dfrac{d}{2}$ from the central maximum.

The path difference $\Delta x$ at this point is given by:
$\Delta x = \dfrac{yd}{D}$

Substituting $y = \dfrac{d}{2}$ and $D = 10d$:
$\Delta x = \dfrac{\left(\dfrac{d}{2}\right)d}{10d} = \dfrac{d^2}{20d} = \dfrac{d}{20}$

Given that the distance between the slits is $d = 5\lambda$, we have:
$\Delta x = \dfrac{5\lambda}{20} = \dfrac{\lambda}{4}$

The corresponding phase difference $\Delta \phi$ is:
$\Delta \phi = \dfrac{2\pi}{\lambda} \Delta x = \dfrac{2\pi}{\lambda} \left(\dfrac{\lambda}{4}\right) = \dfrac{\pi}{2}$

The intensity at this point is given by:
$I = I_0 \cos^2\left(\dfrac{\Delta \phi}{2}\right)$

Substituting $\Delta \phi = \dfrac{\pi}{2}$:
$I = I_0 \cos^2\left(\dfrac{\pi}{4}\right) = I_0 \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{I_0}{2}$

Answer: $\dfrac{I_0}{2}$