Question 1113083
The probability the first ball picked is white is:     P(ball 1 is white) = {{{ n/(2n) = 1/2 }}}
The probability the 2nd is also white:      P(ball 2 is also white) = {{{ (n-1)/(2n-1)  }}}
The probability the 3rd is also white:      P(ball 3 is also white) = {{{ (n-2)/(2n-2)  }}}

P(all 3 picks are white) = {{{ (1/2)*((n-1)/(2n-1)*(n-2)/(2n-2)) =  (1/12) }}} 

Cross multiplying and then simplifying with all terms moved over to the left:

   {{{ (n^2 - 6n + 5) = 0 }}}  
   {{{ (n-5)(n-1) = 0 }}}

Potential solutions are: 1 and 5  (1 and 5 are solutions to the above equation, but might not be solutions to the problem).  

Discard n=1 because that solution is too small to draw 3 balls.
—
Ans:  {{{ highlight(n = 5) }}}
—

Check:   (1/2)*(4/9)*(3/8) = (12/144) = 1/12 (ok)
—
Edited to remove extra factor of n, which was a carry over from my paper solution where I didn't reduce n/(2n) to 1/2.