Answer
${\bf 86.7}\;\rm mT$
Work Step by Step
This problem is similar to the previous one, and we can use the same given graph of Figure P32.68 to make things clear. But we need to change the length of the wire to 12 cm, and the springs' constant to 1.3 N/m.
The force exerted by the springs on the current wire is equal to the magnetic force exerted on the current wire since it is perpendicular to the magnetic field and 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, $F_{\rm springs}=2k\Delta x$,
$$ 2k\Delta x=BIL $$
Hence,
$$\Delta x=\dfrac{BL}{2k}I\tag 1$$
It is now similar to a straight-line formula of $y=mx+b$ where
$\Delta x=y$,
$x=I$,
$m={\rm Slope}=\dfrac{BL}{2k}$,
and $b=0$
Hence,
$$B=\dfrac{2k\;{\rm Slope}}{L}\tag 2$$
So we have to find the slope of this straight line equation above so we can find $B$.
Now we need to draw the graph by plugging the given dots and then draw the best-fit line, as seen below.
From the graph below, we can see that the slope is $4\;\rm mm/A=0.004\;m/A$
Plug into (2),
$$B=\dfrac{2k(0.04)}{L} =\dfrac{2(1.3)(0.004)}{0.12} $$
$$B=\color{red}{\bf 86.7}\;\rm mT$$
