Two light sources having intensities of 8 cd and 12 cd are...

The illuminance $E$ produced by a point source of luminous intensity $I$ at a distance $r$ is given by $E = \dfrac{I}{r^2}$.

Since the two sources of $8 \text{ cd}$ and $12 \text{ cd}$ are placed on the same side of the photometer screen at the same distance, their combined illuminance is the sum of their individual illuminances:

$E_1 = \dfrac{I_1}{r_1^2} + \dfrac{I_2}{r_1^2} = \dfrac{8}{40^2} + \dfrac{12}{40^2} = \dfrac{20}{40^2}$

Let the $80 \text{ cd}$ source be placed at a distance $r_2$ on the opposite side to balance the illuminance. The illuminance produced by this source is:

$E_2 = \dfrac{I_3}{r_2^2} = \dfrac{80}{r_2^2}$

For the photometer to be balanced, $E_1 = E_2$:

$\dfrac{20}{40^2} = \dfrac{80}{r_2^2}$

$r_2^2 = \dfrac{80 \times 40^2}{20}$

$r_2^2 = 4 \times 1600 = 6400$

$r_2 = 80 \text{ cm}$