let 'x' be the length= width of the square picture, x>0

the total area of the picture and the frame: (x+3)^2 = x^2 + 6x + 9

let A be the area of the family crest that is at most two-thirds the total area of the picture and frame

(2/3)*(x^2 + 6x + 9) >= A

2x^2 + 12x + 18 - 3A >= 0

by solving the quadratic equation we find the roots

x1 = -3 - (6A)^(0.5)

x2 = -3 + (6A)^(0.5)

since x>0 the possible length of one side of the frame: x+3 would satisfy the inequality

x+3 >= -3 + (6A)^(0.5)+3

x+3 >= (6A)^(0.5) > 2.44*A

the side of the frame is at least 2.44 of the squared root of the area of the family crest.