In an ac circuit, an inductor, a capacitor and a resistor are connected in series with $$X_L = R = X_C$$. Impedance of this circuit is:

We have an a.c. circuit in which a resistor of resistance $$R$$, an inductor offering inductive reactance $$X_L$$ and a capacitor offering capacitive reactance $$X_C$$ are all connected in series.

The impedance of any series $$R$$-$$L$$-$$C$$ circuit is defined by the formula

$$Z \;=\; \sqrt{\,R^{2} \;+\; \bigl(X_L - X_C\bigr)^{2}}.$$

This expression comes directly from phasor analysis: the resistor contributes a real component $$R$$, while the inductor and capacitor contribute imaginary components $$+jX_L$$ and $$-jX_C$$ respectively; their algebraic sum in the imaginary direction is $$j\,(X_L - X_C)$$. Taking the magnitude of the complex quantity $$R + j\,(X_L - X_C)$$ gives the above square-root relation.

Now the question states that the magnitudes of all three quantities are equal, that is

$$X_L \;=\; R \;=\; X_C.$$

Substituting $$X_L = X_C$$ straight into the impedance formula, we obtain

$$Z = \sqrt{\,R^{2} + \bigl(X_L - X_C\bigr)^{2}} = \sqrt{\,R^{2} + \bigl(X_L - X_L\bigr)^{2}} = \sqrt{\,R^{2} + 0^{2}} = \sqrt{\,R^{2}} = R.$$

Thus the magnitude of the impedance of the circuit reduces exactly to the resistance itself.

Hence, the correct answer is Option B.