#### Answer

$20$ Amps

#### Work Step by Step

29.72
Using the right-hand rule, we see that because the current is downward (-$\hat j$ direction) and the magnetic field is going out of the page (+$\hat k$ direction), the magnetic field must be going to the left (-$\hat j \times \hat k = - \hat i$). Thus, the magnetic force pushes against the force the spring provides (remember that a spring's force is in the direction of its displacement). Thus,
$\Sigma\vec{F}_x = 2\vec{F}_s - \vec{F}_B = 0$
Remember, there are two springs in the diagram so a coefficient of 2 goes before $\vec{F}_s$
$2\vec{F}_s = \vec{F}_B$
$2kx = I\vec{L} \times \vec{B} = I\vec{L}\vec{B}\sin(90) = I\vec{L}\vec{B}$
$\displaystyle I = \frac{2kx}{\vec{L}\vec{B}} = \frac{2(10)(0.010)}{(0.020)(0.5)} = 20$ Amps