Explanation

(d) Resistance of original wire is
$R = \rho\frac{l}{A}$
$\rho$ ,being the specific resistance of wire. When the wire is cut in two equal halves then resistance becomes
$R^{'} = \frac{\rho l/2}{A} = \frac{R}{2}$
Thus, the net resistance of parallel combination of two halves is given by
$R_{net} = \frac{R^{'} \times R^{'}}{R^{'} + R^{'}}$
$\ \ \ \ \ \ \ \ \ \ = \frac{R^{'}}{2} = \frac{R}{2 \times 2} = \frac{6}{4} = 1.5\Omega$