#### Answer

a). Same
b). One-quarter the current

#### Work Step by Step

a). $R=p\frac{L}{A}$, $p$ being the resistivity,
So, when length is 2L, and area is 2A,
there is no change in R, hence current remains same as $I=\frac{V}{R}$.
b). When length is L and area is $\frac{A}{4}$, R becomes 4 times, hence I becomes $\frac{I}{4}$, so it becomes one-quarter the current.