Let $u$ be the object distance and $v$ be the image distance. Let $f$ be the focal length of the mirror.

The magnification ($m$) is given by $m = \frac{v}{u}$.

The question states a "three times magnified real image". For a real image formed by a mirror, the magnification is negative. Thus, $m = -3$.

From the magnification formula, $\frac{v}{u} = -3$, which implies $v = -3u$.

For a concave mirror forming a real image, both the object and the image are in front of the mirror, meaning $u$ and $v$ are negative according to the Cartesian sign convention. However, if $u$ is negative, $v = -3u$ implies $v$ is positive, which corresponds to a virtual image. This indicates a potential inconsistency in applying standard sign conventions directly.

A common interpretation assumes the magnitude of magnification is 3, i.e., $|m|=3$, and the image is real. This means $|v| = 3|u|$. Since it's a real image formed by a concave mirror, $u < 0$ and $v < 0$. Thus, we can set $u = -x$ and $v = -3x$ for some positive value $x = |u|$.

The distance between the object and the image is given as $40 \text{ cm}$. This separation is $|u - v|$.

Since $x$ represents a distance magnitude, $x > 0$. Therefore, $2x = 40$, which gives $x = 20 \text{ cm}$.

So, the object distance is $u = -x = -20 \text{ cm}$, and the image distance is $v = -3x = -60 \text{ cm}$.

Now, we use the mirror formula: $ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} $ $ \frac{1}{f} = \frac{1}{-20 \text{ cm}} + \frac{1}{-60 \text{ cm}} $ $ \frac{1}{f} = -\frac{1}{20} - \frac{1}{60} $ $ \frac{1}{f} = \frac{-3 - 1}{60} = \frac{-4}{60} = -\frac{1}{15} $ $ f = -15 \text{ cm} $ This result corresponds to Option 4. However, based on the provided correct answer, the focal length is $-20$ cm.