The brightness of the bulb depends on the RMS current flowing through it ($P = I^2R$). The current in the series circuit is given by $I = V/Z$, where $Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (2\pi f L)^2}$. To decrease the brightness, the current must decrease, which implies the impedance $Z$ (and thus the inductive reactance $X_L$) must increase.

• Option 1: Decreasing frequency $f$ decreases $X_L$, which increases $I$ and brightness.
• Option 2: Reducing the number of turns $N$ decreases self-inductance $L$ (since $L \propto N^2$ for a solenoid), which decreases $X_L$ and increases brightness .
• Option 3: Adding a capacitor where $X_C = X_L$ brings the circuit to resonance, minimizing impedance ($Z=R$) and maximizing current and brightness.
• Option 4: Inserting an iron rod increases the relative permeability $\mu_r$ of the core material. Since $L = \mu_r \mu_0 n^2 Al$, the inductance $L$ increases significantly . This increases $X_L$ and $Z$, thereby decreasing the current and the bulb's brightness.