Answer
$1.42\times 10^{-3}T$
Work Step by Step
We know that
$B_{solenoid}=\mu_{\circ}nI_s$
$\implies B_{solenoid}=(4\pi \times 10^{-7}T.m/A)(2200turns/m)(0.50A)=1.38\times 10^{-3}T$
and $B_{straight \space wire}=\frac{\mu_{\circ}I}{2\pi r}$
$\implies B_{straight\space wire}=\frac{(4\pi\times 10^{-7}T.m/A)(13A)}{2\pi(0.75\times 10^{-2}m)}$
$B_{straight \space wire}=0.346\times 10^{-3}T$
Now $B_{net}=\sqrt{(B_{solenoid})^2+(B_{straight \space wire})^2}$
We plug in the known values to obtain:
$B_{net}=\sqrt{(1.38\times 10^{-3}T)^2+(0.346\times 10^{-3}T)^2}$
$B_{net}=1.42\times 10^{-3}T$