Answer
As $R_A$ decreases, the difference between $R'$ and $R$ decreases.
Work Step by Step
The total potential difference across $R$ and the ammeter is equal to $V'$
$R' = \frac{V'}{i}$ where $R' = R+R_A$
As $R_A$ decreases, then clearly $R+R_A$ gets closer to $R$
Therefore, as $R_A$ decreases, the difference between $R'$ and $R$ decreases.
Note that $R' = R$ when $R_A = 0$
