Refraction takes place at a concave spherical boundary $($of radius of curvature $R)$ separating glass - air medium. For the image to be real, the object which is placed in glass at a distance $u$ $({\mu }_{glass}=\frac{3}{2})$

The correct option is A $u>3R$
Applying the lens maker formula for spherical concave boundary, we have

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

Where, Refractive index of air, ${\mu }_2=1$

Refractive index of glass, ${\mu }_1=\frac{3}{2}$

Radius of curvature of the surface$=R$

Object distance, $-u$ and Image distance, $v$
(Here, we consider positive sign for right to left direction)
Putting the known value in lens maker formula we get,

$\frac{1}{v}-\frac{1.5}{(-u)}=\frac{1-\left(1.5\right)}{-R}$

$\Rightarrow \frac{1}{v}+\frac{3}{2u}=\frac{1}{2R}$

$\Rightarrow \frac{1}{v}=\frac{1}{2R}-\frac{3}{2u}$

For $v$ to be positive, $\frac{1}{2R}>\frac{3}{2u}$

$\therefore u>3R$

Hence, object distance should be greater than three times of the radius of curvature of refractive surface.