A point source emitting light uniformly in all directions is...

The illuminance at a point directly below the source is given by $E_{0} = \dfrac{I}{h^{2}}$.

Given $E_{0} = 15 \text{ lux}$ and $h = 60 \text{ cm} = 0.6 \text{ m}$.

The illuminance at a point on the table-top at a distance $x$ from the first point is $E = \dfrac{I \cos \theta}{r^{2}}$, where $r = \sqrt{h^{2} + x^{2}}$ and $\cos \theta = \dfrac{h}{r}$.

Substituting $I = E_{0} h^{2}$ and $\cos \theta$, we get:

$E = \dfrac{E_{0} h^{2}}{r^{2}} \times \dfrac{h}{r} = E_{0} \left( \dfrac{h}{r} \right)^{3}$

Given $x = 80 \text{ cm} = 0.8 \text{ m}$, the distance $r$ from the source to the point is:

$r = \sqrt{0.6^{2} + 0.8^{2}} = 1.0 \text{ m}$

Therefore, the illuminance at this point is:

$E = 15 \times \left( \dfrac{0.6}{1.0} \right)^{3} = 15 \times 0.216 = 3.24 \text{ lux}$