Answer
$$ b=\sqrt 3.$$
Work Step by Step
$$
\int_{0}^{b} \frac{d x}{1+x^{2}}=\frac{\pi}{3} \Longrightarrow \tan^{-1}
x|_0^b=\frac{\pi}{3}\\ \tan^{-1}
b- \tan^{-1}0=\frac{\pi}{3}\Longrightarrow \tan^{-1}
b-0=\frac{\pi}{3} \\\Longrightarrow
b=\sqrt 3.
$$
