Answer
$8.94\times 10^{-6}T;116^{\circ}$
Work Step by Step
We know that
$\vec B=\vec B_1+\vec B_2$
$\implies \vec B=(1.2\times 10^{-5}T)\hat x+(8\times 10^{-6}T)(-\hat x)+(8\times 10^{-6}T)(\hat y)$
$\implies \vec B=\sqrt{B_x^2+B_y^2}$
Now the magnitude is given as
$B=\sqrt{(0.4\times 10^{-5}T)^2+(8\times 10^{-6}T)^2}$
$\implies B=8.94\times 10^{-6}T$
Now the direction can be determined as
$\theta=tan^{-1}(\frac{B_y}{B_x})$
$\implies \theta=tan^{-1}(\frac{8\times 10^{-6}T}{0.4\times 10^{-6}T})=63.4^{\circ}$
Now $\theta^{\circ}=180^{\circ}-63.4^{\circ}=116^{\circ}$
