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 cylinder 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 cylindrical surface inside our hollow sphere, then $A=2\pi rL$
$$ \boxed{E =\dfrac{I}{2\pi \sigma L r }}$$
$$\color{blue}{\bf [b]}$$
At the inner surface, when $r=a$
$$ E_{\rm inner} =\dfrac{I}{2\pi \sigma_{\rm iron} L a} $$
Plug the known;
$$ E_{\rm inner} =\dfrac{25}{2\pi (1\times 10^7)(0.1)(0.01)} $$
$$ E_{\rm inner} =\color{red}{\bf 4\times 10^{-4}}\;\rm V/m$$
At the outer surface, when $r=b$
$$ E_{\rm outer } =\dfrac{I}{2\pi \sigma_{\rm iron} L b } $$
Plug the known;
$$ E_{\rm outer } =\dfrac{25}{2\pi (1\times 10^7)(0.1)(0.025) } $$
$$ E_{\rm outer } =\color{red}{\bf 1.59\times 10^{-4}}\;\rm V/m$$
