When positioned at 80 cm and 20 cm from a photometer screen...

The initial illuminance of the electric lamp and the candle are equal on the photometer screen.

$\dfrac{I_1}{r_1^2} = \dfrac{I_2}{r_2^2}$

$\dfrac{I_1}{80^2} = \dfrac{I_2}{20^2} \Rightarrow I_1 = 16 I_2$

When the lamp is covered with paper, its new luminous intensity becomes $I_1' = 0.49 I_1$.

To restore balanced intensities, let the new distance of the lamp be $r_1'$. The illuminance must again be equal:

$\dfrac{I_1'}{(r_1')^2} = \dfrac{I_2}{r_2^2}$

$\dfrac{0.49 I_1}{(r_1')^2} = \dfrac{I_2}{20^2}$

Substituting $I_1 = 16 I_2$:

$\dfrac{0.49 \times 16 I_2}{(r_1')^2} = \dfrac{I_2}{400}$

$(r_1')^2 = 0.49 \times 16 \times 400 = 3136$

$r_1' = 56 \text{ cm}$

The distance by which the lamp must be shifted is $\Delta r = 80 - 56 = 24 \text{ cm}$.