A straight horizontal wire has a mass of 25 g in each metre of its length…

Step 1. The direction. Let east be $\hat{i}$, north be $\hat{j}$ and up be $\hat{k}$. The field is along $\hat{j}$. A current to the east has $\vec{L}$ along $\hat{i}$, and $\hat{i} \times \hat{j} = \hat{k}$, which is up. So the current flows east.

Step 2. The current. For each metre of wire, $BIL = mg$ with $L = 1$ m and $m = 25 \times 10^{-3}$ kg. We want $I$, so $I = \dfrac{mg}{BL} = \dfrac{25 \times 10^{-3}\ \text{kg} \times 9.8\ \text{m/s}^2}{0.080\ \text{T} \times 1\ \text{m}} = 3.06\ \text{A}$.

Step 3. Twice the current. The magnetic force is now $2mg$ and the weight is $mg$. The net force is $mg$, upward, so $a = \dfrac{mg}{m} = g = 9.8\ \text{m/s}^2$.