Answer:
The distance is  [tex]x =0.291 \ m[/tex]
Explanation:
From the question we are told that
   The width of the box is [tex]b = 10 \ cm = \frac{10}{100} = 0.10 \ m[/tex]
   The gap is of length [tex]l = 1\ m[/tex]
  The natural length of the first spring is [tex]y = 80 \ cm = \frac{80 }{100} = 0.8 \ m[/tex]
   The spring constant of the first spring is  [tex]k_1 = 200 N/m[/tex]
   The second spring has a natural length of [tex]z = 90 \ cm = \frac{90}{100} = 0.9 \ m[/tex]
   The spring constant of the first spring is  [tex]k_2 = 350 \ N/m[/tex]
Let the distance from the center of the box to the left edge be x
   So at equilibrium
The force applied by the first spring is
   [tex]F_1 = k_1 * (0.8 -x)[/tex]
and
The force applied by the second spring is
    [tex]F_2 = k_2 * [ 0.9 - (0.9 -x)][/tex]
Now at equilibrium
    [tex]F_1 = F_2[/tex]
So
   [tex]k_1 * (0.8 -x) = k_2 * [ 0.9 - (0.9 -x)][/tex]
  substituting values
   [tex]200 * (0.8 -x) = 350 * [ 0.9 - (0.9 -x)][/tex]
    =>  [tex]160 -200x) = 350x[/tex]
    =>  [tex]160 =550x[/tex]
     =>  [tex]x =0.291 \ m[/tex]
