As illustrated in the figure, a small diffused plane source S is...

For a small diffused plane source, the luminous intensity in a direction making an angle $\theta$ with the normal is given by Lambert's cosine law:

$I(\theta) = I_0 \cos \theta$

The illuminance $E$ at a point on a surface is given by:

$E = \dfrac{I(\theta)}{r^2} \cos \phi$

where $r$ is the distance from the source to the point, and $\phi$ is the angle of incidence. Since the source plane is parallel to the table-top, the angle of emission $\theta$ is equal to the angle of incidence $\phi$. Thus:

$E = \dfrac{I_0 \cos^2 \theta}{r^2}$

At point $A$, the point is directly beneath the source, so $\theta = 0^\circ$ and $r_A = 2.4 \text{ m}$.

$E_A = \dfrac{I_0}{(2.4)^2} = 25 \Rightarrow I_0 = 25 \times 5.76 = 144 \text{ cd}$

For point $B$, the distance from the source is:

$r_B = \sqrt{(2.4)^2 + (1.8)^2} = \sqrt{5.76 + 3.24} = \sqrt{9.00} = 3.0 \text{ m}$

The angle $\theta$ for point $B$ is given by:

$\cos \theta = \dfrac{2.4}{3.0} = 0.8$

The illuminance at $B$ is:

$E_B = \dfrac{I_0 \cos^2 \theta}{r_B^2} = \dfrac{144 \times (0.8)^2}{(3.0)^2} = \dfrac{144 \times 0.64}{9} = 16 \times 0.64 = 10.24 \text{ lux}$