Maths problem, how to do, thanks?

(a)
Let height of a larger cylinder = H
base area of a larger cylinder = πR² 
Then, total vol. of the 2 larger cylinders = 2πR²H

base area of a smaller cylinder = πr²
Also, base area of a smaller cylinder = ¼ πR²
∴ ¼ πR² = πr²
=> R² = 4r²
=> R= 2r - - - - - - (#)

total vol. of the 20 smaller cylinders=20πr²(6)=120πr²

As the 2 larger cylinders are recast in 20 smaller cylinders, 
2πR²H = 120πr²
R²H = 60r²
∴ H = 60(r/R)²
　　= 60(1/2)²　　[ From (#) ]
　　= 15 cm

(b) 
ratio of curved surface area of a larger cylinder to
curved surface area of a smaller cylinder
= 2πR(15) : 2πr(6)
= R(15) : r(6)
= 5/2(R/r)
= (5/2)*(2)　　[ From (#) ]
= 5:1

2πR^2H = 20(πr^2h) = 120πr^2 (∵ h = 6cm)
πR^2 = 4πr^2 (∴ R = 2r)∴ 8πr^2 H = 120πr^2∴ H = 15(cm)我不確定 curved surface 是指側面積還是包含上下底.若只計側面積(個人猜測是問這個), 則 (2πRH)/(2πrh) = (R/r)(H/h) = 2(15/6) = 5若含上下底, 則表面積比值與r有關:(2πR^2+2πRH)/(2πr^2+2πrh)   = (8πr^2+60πr)/(2πr^2+12πr)   = (4r+30)/(r+6)
是底面積倍數與側表面積倍數之加權平均.