F.3 maths question~plzz help!!

(a)
Let the radius of small cylinder be = xR
Let the height of small cylinder be = xH
x is the conversion factor 0 < x < 1, xR must be smaller than R
volume of large cylinder = 27 times volume of smaller cylinder
pi (R^2) (H) = 27 pi [(xR)^2] (xH)
(R^2) (H) = 27 (x^2)(R)^2 x(H)
(R^2) (H) = 27 (x^3)(R^2) (H)
1 = 27 (x^3)
x^3 = 1/27
x = 1/3
The base radius of the small candle = xR = (1/3) R

(b)
Surface area of large cylinder = 2(pi)R^2 + 2 (pi)RH

Total Surface area of 27 small cylinders = 27 [2(pi)(R/3)^2 + 2 (pi)(R/3)(H/3)]
= 27/9 [2(pi)(R)^2 + 2 (pi)(R)(H)]
= 3[2(pi)(R)^2 + 2 (pi)(R)(H)]
Total Surface area of 27 small cylinders = 3 times Surface area of large cylinder
Let S.A. of large cylinder be A and S.A. of 27 small cylinders be 3A

% increase = (S.A. of 27 small cylinder – S.A. of large cylinder)/ S.A. of large cylinder x 100%
= [(3A – A)/A] x 100% = [2A/A] x 100% = 200%

The percentage increase in the total surface area of the candles after the process is 200%