Three wires each of length L , cross-sectional area A and...

The resistance of a single wire of length $L$, cross-sectional area $A$, and resistivity $\rho$ is given by $R = \rho \dfrac{L}{A}$.

Between points $X$ and $Y$, there are two such wires connected in parallel. The equivalent resistance $R_{XY}$ is:
$\dfrac{1}{R_{XY}} = \dfrac{1}{R} + \dfrac{1}{R} = \dfrac{2}{R} \Rightarrow R_{XY} = \dfrac{R}{2} = \dfrac{\rho L}{2A}$

Between points $Y$ and $Z$, there is one such wire. Its resistance is $R_{YZ} = R = \dfrac{\rho L}{A}$.

The total resistance between $X$ and $Z$ is the series combination of $R_{XY}$ and $R_{YZ}$:
$R_{XZ} = R_{XY} + R_{YZ} = \dfrac{\rho L}{2A} + \dfrac{\rho L}{A} = \dfrac{3\rho L}{2A}$

The new wire has length $L_1$, area $A_1$, and the same resistivity $\rho$. Its resistance is $R_{new} = \rho \dfrac{L_1}{A_1}$.

Since the new wire replaces the combination without changing the resistance, we have:
$R_{new} = R_{XZ}$
$\rho \dfrac{L_1}{A_1} = \dfrac{3\rho L}{2A}$
$\dfrac{L_1}{A_1} = \dfrac{3L}{2A}$

Checking the given options:
(a) $L_1 = 3L$, $A_1 = 2A \Rightarrow \dfrac{L_1}{A_1} = \dfrac{3L}{2A}$ (Matches)
(b) $L_1 = L$, $A_1 = A \Rightarrow \dfrac{L_1}{A_1} = \dfrac{L}{A}$ (Incorrect)
(c) $L_1 = 2L$, $A_1 = 3A \Rightarrow \dfrac{L_1}{A_1} = \dfrac{2L}{3A}$ (Incorrect)
(d) $L_1 = 2L$, $A_1 = 2A \Rightarrow \dfrac{L_1}{A_1} = \dfrac{L}{A}$ (Incorrect)

Answer: $L_1 = 3L$ and $A_1 = 2A$