(a) The magnitude of the electric field is given by:
[tex]E=\frac{100}{r}[/tex]
where r is the distance from the centre in metres. Therefore, if we substitute
[tex]r=9 cm=0.09 m[/tex]
we can find the magnitude of the electric field at a distance of 9 cm:
[tex]E=\frac{100}{r} =\frac{100}{0.09} =1111.1 N/C[/tex]
(b) The electric field direction is radial at every point of the space (in fact, the problem gives us the direction of the electric field vector: e = (100/r)r̂ n/c, where r̂ indicates that the field is radial). Therefore, the electric field is perpendicular to the surface of the concentric sphere of radius r=18 cm=0.18 m, so the flux of the electric field through the surface of the sphere is simply equal to the product between the electric field strength calculated at the surface of the sphere and the surface of the sphere:
[tex]\Phi = EA=(\frac{100}{r})(4 \pi r^2)=(\frac{100}{0.18})(4 \pi (0.18)^2)=226.1 Nm^2/C[/tex]
