Answer
${\bf 2.0}\;\rm A$
Work Step by Step
The force exerted by the springs is equal to the magnetic force exerted on the current wire since it is now at rest.
Thus,
$$\sum F_x=F_{\rm springs}-F_{\rm magnetic}=0$$
$$ F_{\rm springs}=F_{\rm magnetic} $$
and since the two springs are identical, $F_{\rm springs}=2k\Delta x$,
$$ 2k\Delta x=BIL $$
Solving for $I$ to find the current,
$$I=\dfrac{2k\Delta x}{BL}$$
Plug the given;
$$I=\dfrac{2(10)(0.01) }{(0.50)(0.20)}$$
$$I=\color{red}{\bf 2.0}\;\rm A$$
