% error..

percentage error in measuring surface area is x%

CASE I: x is small

Then % error in measuring length is x/2%
So % error in volume is 3x/2%

CASE II: x is not small

Let the % error in measuring length be a

Then if the length is measured as l, the actual length is l(1±a/100)

Let actual length be l(1+a/100) then surface area = l²(1+a/50+a²/10000)

So %error = (a/50+a²/10000)*100 = 2a+a²/100 = x ...(i)

volume = l_act³ = l³(1+3a/100 + 3a²/10000 + a³/10⁶)

So % error = (3a/100 + 3a²/10000 + a³/10⁶)*100 = 3a + 3a²/100 + a³/10000

u can use the value of a obtained in (i)