Use the resistance formula

For a wire, $R = \rho \frac{L}{A}$ where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area.

Apply conservation of volume

When the wire is melted and redrawn, its material and hence its volume remain the same: $A L = A' L'$

Given the new length is one-fourth of the original: $L' = \frac{L}{4}$

So, $A' \cdot \frac{L}{4} = A L$ $A' = 4A$

Find the new resistance

The new resistance is $R' = \rho \frac{L'}{A'} = \rho \frac{L/4}{4A} = \rho \frac{L}{16A} = \frac{R}{16}$

Substitute the original resistance

$R' = \frac{160}{16} = 10~\Omega$

Match with the options

Therefore, the new resistance is: $\boxed{10~\Omega}$

So the correct option is D.