As we know,
$
R =\frac{ \rho L }{ A } .
$
If you stretch the wire till it's radius is halved the previous value, then it's length will increase 4 times to keep the volume constant.
So, new values are,
$
\begin{array}{l}
L _2=4 L \\
A _2=\frac{ A }{4}
\end{array}
$
where, $L$ and $A$ are the values of original length and area
Resistivity remains the same as it depends upon the material of the wire.
So, the new resistance will be,
$
\begin{array}{l}
R _1=\frac{\rho L }{ A } \\
R _1=\frac{\rho 4 L }{ A / 2} \\
R _1=\frac{16 \rho L }{ A } \\
R _1=16 R
\end{array}
$