I think that this is all that Heureka did. It is all that I would normally do.
if the ratio if a 1 dimensional measurement is $$a:b$$ (the radius is a 1 dimensional measurement)
then the ration of the 2 dimensional measurements (surface area) is just $$a^2:b^2$$
and the ratio of the 3 dimensional measurement (volume) is just $$a^3:b^3$$
so if ratio of radius is 1:2
then ratio of SA is 1:4
and ratio of Volume is 1:8
(i) ratio among their surface areas is 4:1.
(ii) the ratio among their volumes is 8:1.
(i) ratio among their surface areas is 4:1.
$$r_a=2r_b$$
$$\begin{array}{rcl}
Surface_A &=& 4\pi(2r_B)^2 \\
Surface_B &=& 4\pi(r_B)^2 \\
\hline
\\
\frac{Surface_A}{Surface_B} &=& \frac{4\pi(2r_B)^2}{4\pi(r_B)^2
} \\\\
\frac{Surface_A}{Surface_B} &=& \frac{4\pi (2)^2 (r_B)^2}{4\pi(r_B)^2
}=2^2 = 4= \frac{4}{1}
\\\\
\end{array}$$
(ii) the ratio among their volumes is 8:1.
$$\begin{array}{rcl}
V_A &=& \frac{4}{3}\pi(2r_B)^3 \\\\
V_B &=& \frac{4}{3}\pi(r_B)^3 \\\\
\hline
\\
\frac{V_A}{V_B} &=& \frac{\frac{4}{3}\pi(2r_B)^3}{\frac{4}{3}\pi(r_B)^3
} \\\\
\frac{V_A}{V_B} &=& \frac{\frac{4}{3}\pi(2)^3(r_B)^3}{\frac{4}{3}\pi(r_B)^3
} = 2^3 = 8 = \frac{8}{1} \\\\
\end{array}$$
I think that this is all that Heureka did. It is all that I would normally do.
if the ratio if a 1 dimensional measurement is $$a:b$$ (the radius is a 1 dimensional measurement)
then the ration of the 2 dimensional measurements (surface area) is just $$a^2:b^2$$
and the ratio of the 3 dimensional measurement (volume) is just $$a^3:b^3$$
so if ratio of radius is 1:2
then ratio of SA is 1:4
and ratio of Volume is 1:8