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I'm assuming that the area  of the triangle   is    (75/4) * sqrt (3)

We can solve for the side length, S,  of the triangle as follows

(75/ 4)sqrt (3)  =  (1/2) S^2  sin (60°)

(75/4)sqrt (3)  = (1/2) S^2 * sqrt (3) / 2

(75/4)  = (1/4)S^2

75  = S^2

sqrt (75)  = S

And  using the Law of Cosines we can find the  radius, R, of the  circle

75 =  2R^2  - 2R^2  cos (120°)

75 = 2R^2  + R^2

75  = 3 R^2

25 = R^2

5  = R

And the side of the square is twice  this  =  10

So.....the perimeter of the  square = 4 * 10   =   40 units

The area of the inscribed equilateral triangle is 75/4*sqrt(3)​​. Find the perimeter of circumscribing square to the nearest whole number.

The ratio of the area of an equilateral triangle to its circumcircle's area is:   1.299 : pi  or  0.413496

Area of the triangle is        75/4*sqrt(3) = 32.47595 u²

Area of the circle is            32.47595 / 0.413496 = 78.54 u²

Raduis of the circle is        r = sqrt( 78.54 / pi ) = 5

Perimeter of a square is     P = 8r = 40 units