1. Magnetic field due to a long straight wire

   The magnetic field at distance $r$ from a long straight wire carrying current $I$ is $B=\frac{\mu_0 I}{2\pi r}$

2. Understand the geometry

   The two wires $X$ and $Y$ are separated by $6\,\text{cm}$.

   Point $P$ is at a distance $4\,\text{cm}$ from wire $Y$.

   From the standard figure for this question, $P$ lies on the line joining the wires, on the side of wire $Y$ away from wire $X$. Therefore, $XP = XY + YP = 6+4=10\,\text{cm}$

   So, $r_Y=4\,\text{cm}=0.04\,\text{m}, \qquad r_X=10\,\text{cm}=0.10\,\text{m}$

3. Field at $P$ due to wire $Y$

   Current in wire $Y$ is $4\,\text{A}$.

   =\frac{4\pi\times 10^{-7}\times 4}{2\pi\times 0.04}$$ $$B_Y=\frac{16\pi\times 10^{-7}}{0.08\pi}=2\times 10^{-5}\,\text{T}$$

4. Field at $P$ due to wire $X$

   Current in wire $X$ is $5\,\text{A}$.

   =\frac{4\pi\times 10^{-7}\times 5}{2\pi\times 0.10}$$ $$B_X=\frac{20\pi\times 10^{-7}}{0.20\pi}=1\times 10^{-5}\,\text{T}$$

5. Direction of the fields

   The currents are in opposite directions. At point $P$ (outside the two wires), by the right-hand rule, the magnetic fields due to the two wires are in opposite directions.

   Hence the resultant magnitude is $B_{\text{net}}=|B_Y-B_X|=|2-1|\times 10^{-5}$ $B_{\text{net}}=1\times 10^{-5}\,\text{T}$

6. Find $x$

   Given $B_{\text{net}}=x\times 10^{-5}\,\text{T}$ so, $x=1$

Final Answer: $\boxed{1}$