The intensity $I$ of light from a light bulb varies inversely as the square of the distance $d$ from the light bulb. Suppose $l$ is $360 \mathrm{~W} / \mathrm{m}^{2}$ when the distance is $3 \mathrm{~m}$. How much further would it be to a point where the intensity is $250 \mathrm{~W} / \mathrm{m}^{2}$ ? It would be $\square$ $\square$ farther. (Simplify your answer. Type an integer or a decimal.)

Step 1 ：Given that the intensity of light from a light bulb varies inversely as the square of the distance from the light bulb, we can use the formula \(I = \frac{k}{d^2}\), where \(I\) is the intensity, \(d\) is the distance, and \(k\) is a constant.

Step 2 ：Given that the intensity \(I_1\) is \(360 \, \mathrm{W/m^2}\) when the distance \(d_1\) is \(3 \, \mathrm{m}\), we can find the value of \(k\) using the formula \(k = I_1 \cdot d_1^2\). Substituting the given values, we get \(k = 360 \cdot 3^2 = 3240\).

Step 3 ：We are asked to find the additional distance to a point where the intensity \(I_2\) is \(250 \, \mathrm{W/m^2}\). We can find the new distance \(d_2\) using the formula \(d_2 = \sqrt{\frac{k}{I_2}}\). Substituting the values, we get \(d_2 = \sqrt{\frac{3240}{250}} \approx 3.6 \, \mathrm{m}\).

Step 4 ：The additional distance would be the difference between the new distance and the original distance, which is \(d_2 - d_1 = 3.6 - 3 = 0.6 \, \mathrm{m}\).

Step 5 ：Final Answer: The additional distance to a point where the intensity of light is \(250 \, \mathrm{W/m^2}\) is \(\boxed{0.6} \, \mathrm{m}\).