Answer
$0.5\;\rm mm$
Work Step by Step
To find the diameter of the wire for this function, we need to find its cross-sectional area which we can find, due to the given, by applying the current density.
$$J=\dfrac{I}{A}$$
where $A=\pi r^2=\pi (D/2)^2=\pi D^2/4$ whereas $D$ is the wire's diameter.
$$J=\dfrac{4I}{\pi D^2}$$
Solving for $D$;
$$D=\sqrt{\dfrac{4I}{\pi J}}$$
Plug the known;
$$D=\sqrt{\dfrac{4(1.0)}{\pi (500\times 10^4)}}$$
$$D=\color{red}{\bf 0.50}\;\rm mm$$
which is a reasonable answer since we know that the fuse wire had to be thin.
