Given, current in a long wire, $I=18 \mathrm{~A}$


magnetic field due to a solenoid,
$$
\begin{aligned}
& \qquad B_1=8 \times 10^{-3} \mathrm{~T} \\
& \text { and } r=0.6 \mathrm{~mm}=0.6 \times 10^{-3} \mathrm{~m} \\
& \text { Radius of solenoid, } R=1 \mathrm{~cm}=1 \times 10^{-2} \mathrm{~m} \\
& \text { Magnetic field due to current carrying wire at a } \\
& \text { point, } P \\
& \qquad \begin{aligned}
B_2 & =\frac{\mu_0}{2 \pi} \cdot \frac{I}{r}=2 \times 10^{-7} \times \frac{18}{0.6 \times 10^{-3}} \\
& =6 \times 10^{-3} \mathrm{~T} \quad \text { (in upward direction) }
\end{aligned}
\end{aligned}
$$

Since, current carrying wire is placed along the axis of solenoid, hence magnetic field produced by wire and magnetic field due to solenoid at point $P$, both are perpendicular to each other, Hence, the magnitude of the resultant magnetic field,
$$
\begin{aligned}
B & =\sqrt{B_1^2+B_2^2} \\
& =\sqrt{\left(8 \times 10^{-3}\right)^2+\left(6 \times 10^{-3}\right)^2} \\
& =10 \times 10^{-3} \mathrm{~T}
\end{aligned}
$$