Given : $\mathit{u}=-15\mathrm{cm},\mathit{f}=-10\mathrm{cm},{\mathit{h}}_{\mathit{o}}=4\mathrm{cm}$
Using the mirror formula
$\frac{1}{\mathit{v}}+\frac{1}{\mathit{u}}=\frac{1}{\mathit{f}}$
$\frac{1}{\mathit{v}}-\frac{1}{15}=\frac{-1}{10}$
$\frac{1}{\mathit{v}}=\frac{-1}{10}+\frac{1}{15}$
$\frac{1}{\mathit{v}}=\frac{-15+10}{150}=\frac{-5}{150}$
$\mathit{v}=-30\mathrm{cm}$
Thus, to obtain a sharp image of the object, the screen should be placed in front of the mirror at a distance of $30\mathrm{cm}$.
Now,

$\mathit{m}=-\left(\frac{-30}{-15}\right)\Rightarrow \mathit{m}=-2$
$-2=\frac{{\mathit{h}}_{\mathit{i}}}{4}\Rightarrow {\mathit{h}}_{\mathit{i}}=-8\mathrm{cm}$
$\therefore$ Height of the image is $8\mathrm{cm}$. (inverted)