Answer
$\frac{A'}{A} = 0.730$
Work Step by Step
Let $A'$ be the new area and let $L'$ be the new length.
Since the volume of the wire remains unchanged, then $~~A'~L' = A~L$
Thus: $A' = \frac{A~L}{L'}$
We can use the new resistance $R'$ to find $\frac{L'}{L}$:
$R' = \frac{30.0~P}{(4.00~i)^2} = \frac{30.0}{16.0}~R$
$\frac{\rho~L'}{A'} = \frac{30.0}{16.0}~\frac{\rho~L}{A}$
$\frac{L'}{AL/L'} = \frac{30.0}{16.0}~\frac{L}{A}$
$\frac{(L')^2}{AL} = \frac{30.0}{16.0}~\frac{L}{A}$
$\frac{(L')^2}{L^2} = \frac{30.0}{16.0}$
$\frac{L'}{L} = \sqrt{\frac{30.0}{16.0}}$
$\frac{L'}{L} = 1.37$
We can find $\frac{A'}{A}$:
$A'~L' = A~L$
$\frac{A'}{A} = \frac{L}{L'}$
$\frac{A'}{A} = \frac{1}{1.37}$
$\frac{A'}{A} = 0.730$
