Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Let's assume that this infinite sheet is formed from an infinite number of parallel closely spaced current-carrying wires, as the author told us.
The current from each wire is coming out of the page toward you, so according to the right-hand rule, the direction of the magnetic field of one wire is counterclockwise.
This means that for a wire in the middle of the sheet, the direction of its magnetic field is pointing toward the left above the sheet and pointing toward the right below the sheet.
The magnetic field between any two wires (in the horizontal direction NOT above the wire or below it, inside the sheet itself) is zero since the magnetic field of the right wire is downward while for the left wire next to it the magnetic field is upward and both cancel each other.
Adding these magnetic fields of the too many wires will give us straight magnetic field lines above the sheet that are pointing toward the left and the same amount of magnetic field lines below the sheet but are pointing toward the right.
See the figures below.
$$\color{blue}{\bf [b]}$$
Applying Ampere's law for a closed path, see the blue box in the last figure below.
$$\oint Bds=\mu_0I_{\rm through}$$
$$\int_{\rm Right} Bds+\int_{\rm Left} Bds+\int_{\rm Up} Bds+\int_{\rm down} Bds=\mu_0I_{\rm through}$$
where for the right and left the integral is zero since $\theta=90^\circ$
$$0+0+BL+BL=\mu_0I_{\rm through}$$
$$2BL=\mu_0I_{\rm through}\tag 1$$
where we know that the linear current density is $J_s$ where $J_s=\dfrac{I_{\rm through}}{L}$, so that $$I_{\rm through}=J_sL$$
Plug into (1),
$$2BL=\mu_0J_sL$$
Thus,
$$\boxed{B=\frac{1}{2}\mu_0J_s}$$
