Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the current is radially outward from the inner surface to the outer surface, the current density at a distance $r$ (where $a\lt r\lt b$) from the center of the sphere is given by
$$J =\dfrac{I}{A}$$
where $J$ is also given by $\sigma E$
Hence,
$$\sigma E =\dfrac{I}{A}$$
Thus the electric field is given by
$$ E =\dfrac{I}{\sigma A}$$
And if we assume that there is a spherical surface inside our hollow sphere, then $A=4\pi r^2$
$$ \boxed{E =\dfrac{I}{4\pi \sigma r^2}}$$
$$\color{blue}{\bf [b]}$$
At the inner surface, when $r=a$
$$ E_{\rm inner} =\dfrac{I}{4\pi \sigma_{\rm copper} a^2} $$
Plug the known;
$$ E_{\rm inner} =\dfrac{25}{4\pi (6\times 10^7)(0.01)^2} $$
$$ E_{\rm inner} =\color{red}{\bf 3.3\times 10^{-4}}\;\rm V/m$$
At the outer surface, when $r=b$
$$ E_{\rm outer } =\dfrac{I}{4\pi \sigma_{\rm copper} b^2} $$
Plug the known;
$$ E_{\rm outer } =\dfrac{25}{4\pi (6\times 10^7)(0.025)^2} $$
$$ E_{\rm outer } =\color{red}{\bf 5.3\times 10^{-5}}\;\rm V/m$$