or $\frac{ Q _{ A }}{ Q _{ B }}=\frac{ R _{ A }}{ R _{ B }}$

As $R_BQ_B$

Above equation represents $B$ is correct option.

From Gauss law the electric field inside a spherical shell is zero so option A is correct.

Now $\sigma_A=\frac{ Q _{ A }}{4 \pi R _A^2}$ and $\sigma_B=\frac{ Q _{ B }}{4 \pi R _{ B }^2}$

Thus $\frac{\sigma_A}{\sigma_B}=\frac{ Q _{ A }}{ Q _{ B }} \times\left(\frac{ R _{ B }}{ R _{ A }}\right)^2=\frac{ R _{ A }}{ R _{ B }} \times\left(\frac{ R _{ B }}{ R _{ A }}\right)^2=\frac{ R _{ B }}{ R _{ A }}$

Thus option C is also correct.

Electric fields on the surface of shell and sphere are

$E _{ A }=\frac{\sigma_{ A }}{\epsilon_0}$

and $E _{ B }=\frac{\sigma_B}{\epsilon_0}$

Thus $\frac{E_A}{E_B}=\frac{\sigma_A}{\sigma_B}<1$

or$E _{ A }< E _{ B }$