Step 1: 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. Step 2: Determine the magnification ($M$). 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. $$M = -3$$ Step 3: Use the magnification formula to relate image distance ($v$) and object distance ($u$). 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. Step 4: Use the mirror formula to solve for the object distance ($u$). 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. The object distance is $\boxed{\text{24 cm}}$.