Given, Focal length of the concave mirror, $f = 4.0 \text{ cm}$ Height of the object, $h_o = 2.5 \text{ mm} = 0.25 \text{ cm}$ Height of the real image, $h_i = -1.0 \text{ cm}$ (since the real image is inverted) Let the object distance be $u$ and the image distance be $v$. The magnification $m$ is given by, $$ \begin{aligned} m &= \frac{h_i}{h_o} = -\frac{v}{u} \end{aligned} $$ Substituting the values, $$ \begin{aligned} \frac{-1.0 \text{ cm}}{0.25 \text{ cm}} &= -\frac{v}{u} \\ -4 &= -\frac{v}{u} \\ \Rightarrow v &= 4u \end{aligned} $$ Now, using the mirror formula, $$ \begin{aligned} \frac{1}{u} + \frac{1}{v} &= \frac{1}{f} \end{aligned} $$ Since it is a concave mirror, $f = -4.0 \text{ cm}$. Substituting $v = 4u$ and $f = -4.0 \text{ cm}$ into the formula, $$ \begin{aligned} \frac{1}{u} + \frac{1}{4u} &= \frac{1}{-4} \\ \frac{4 + 1}{4u} &= -\frac{1}{4} \\ \frac{5}{4u} &= -\frac{1}{4} \\ \Rightarrow 4u &= -20 \\ \Rightarrow u &= -5 \text{ cm} \end{aligned} $$ The negative sign indicates that the object is placed in front of the mirror. Thus, the object should be placed at a distance of $5 \text{ cm}$ from the mirror.