Answer
a) $9$ lumens
b) intensity decreases by one-fourth times
Work Step by Step
a) Using the given formula, with $x=5$, then
$$\begin{aligned}
I&=\frac{225}{x^2}
\\
I&=\frac{225}{5^2}
\\&=
9
.\end{aligned}$$Hence, the intensity, $I$, of the bulb is $9$ lumens.
b) When the distance $x$ doubles, then
$$\begin{aligned}
I_2&=\frac{225}{(2x)^2}
\\&=
\frac{225}{4x^2}
.\end{aligned}$$
Dividing $I_2$ and $I$ results to
$$\begin{aligned}
\frac{I_2}{I}&=\frac{\frac{225}{4x^2}}{\frac{225}{x^2}}
\\&=
\frac{225}{4x^2}\div\frac{225}{x^2}
\\&=
\frac{225}{4x^2}\cdot\frac{x^2}{225}
\\&=
\frac{1}{4}
.\end{aligned}$$Hence, as the distance from the bulb doubles, then the intensity decreases by one-fourth times.
