Calculate the focal length (f) of the concave mirror.

The radius of curvature ($R$) is given as $36 \, \text{cm}$.
For a concave mirror, the focal length is half the radius of curvature:

$$f = \frac{R}{2}$$ $$f = \frac{36 \, \text{cm}}{2}$$ $$f = 18 \, \text{cm}$$

For a concave mirror, the focal length is positive.

The problem states that a real image is formed, which is three times the size of the object.
For a real image formed by a concave mirror, the image is always inverted. Therefore, the magnification is negative.

$$M = -3$$

The magnification formula for mirrors is:

$$M = -\frac{v}{u}$$

Substitute the value of $M$:

$$-3 = -\frac{v}{u}$$ $$v = 3u$$

Since $u$ (object distance) is always positive for a real object, $v$ (image distance) is also positive, which is consistent with a real image formed in front of the mirror.

The mirror formula is:

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$

Substitute $f = 18 \, \text{cm}$ and $v = 3u$ into the mirror formula:

$$\frac{1}{18} = \frac{1}{u} + \frac{1}{3u}$$

To combine the terms on the right side, find a common denominator, which is $3u$:

$$\frac{1}{18} = \frac{3}{3u} + \frac{1}{3u}$$ $$\frac{1}{18} = \frac{3+1}{3u}$$ $$\frac{1}{18} = \frac{4}{3u}$$

Now, cross-multiply to solve for $u$:

$$3u = 4 \times 18$$ $$3u = 72$$ $$u = \frac{72}{3}$$ $$u = 24 \, \text{cm}$$

The object should be placed at a distance of $24 \, \text{cm}$ from the mirror.