The coil of an electric bulb requires 40 W to begin glowing. When...

The power consumed by the bulb is given by $P = \dfrac{V^2}{R}$.

At $V_1 = 220\text{ V}$, the power consumed is $P_1 = 100\text{ W}$.
The light power output at $220\text{ V}$ is:
$L_1 = 0.6 \times (P_1 - 40) = 0.6 \times (100 - 40) = 36\text{ W}$

At $V_2 = 200\text{ V}$, assuming the resistance $R$ remains constant, the power consumed is:
$P_2 = P_1 \times \left(\dfrac{V_2}{V_1}\right)^2 = 100 \times \left(\dfrac{200}{220}\right)^2 = 100 \times \dfrac{100}{121} = \dfrac{10000}{121}\text{ W}$

The light power output at $200\text{ V}$ is:
$L_2 = 0.6 \times \left(P_2 - 40\right) = 0.6 \times \left(\dfrac{10000}{121} - 40\right) = 0.6 \times \left(\dfrac{10000 - 4840}{121}\right) = 0.6 \times \dfrac{5160}{121} = \dfrac{3096}{121}\text{ W}$

The light intensity is directly proportional to the light power output. The percentage drop in light intensity is:
$\text{Percentage drop} = \dfrac{L_1 - L_2}{L_1} \times 100\%$

$\text{Percentage drop} = \dfrac{36 - \dfrac{3096}{121}}{36} \times 100\% = \left(1 - \dfrac{3096}{36 \times 121}\right) \times 100\%$

$\text{Percentage drop} = \left(1 - \dfrac{86}{121}\right) \times 100\% = \dfrac{35}{121} \times 100\% \approx 28.92\%$

Rounding to the nearest integer, the percentage drop is $29\%$.