Solution :

Magnetic field due to a long current carrying wire at distance $$r$$ at point $$P$$ is given by
$$B = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{2i}}{r}$$
So, $$B \propto \frac{1}{r}$$
when $$r$$ is doubled, the magnetic field becomes halved.
i.e., $$B' = \frac{B}{2} \Rightarrow B' = \frac{{0.4}}{2} = 0.2\,T$$