Volume of the solid sphere = \(\frac{4}{3}\)πr³

Volume of twenty seven solid sphere = 27 × \(\frac{4}{3}\)πr³ 

= 36 π r³

(i) New solid iron sphere radius = r’

Volume of this new sphere = \(\frac{4}{3}\)π(r’)³

\(\frac{4}{3}\)π(r’)³ = 36 π r³

(r’)³ = 27r³

r’= 3r

Radius of new sphere will be 3r (thrice the radius of original sphere)

(ii) Surface area of iron sphere of radius r, S =4πr²

Surface area of iron sphere of radius r’= 4π (r’)²

Now

S/S’ = (4πr²)/( 4π (r’)²)

S/S’ = r²/(3r’)² = \(\frac{1}{9}\)

The ratio of S and S’ is 1:9.