To reduce the magnification from $+\frac{1}{2}$ to $+\frac{1}{3}$, the object should be moved further to the spherical mirror.Let the object distance be $u$ and the image distance be $v$. The magnification is given by: magnification $=-\frac{v}{u}$Since the initial magnification is $+\frac{1}{2}$, we have: $+\frac{1}{2}=-\frac{v}{20\,\text{cm}}$Solving for $v$, we get: $v=-10\,\text{cm}$Substituting this value of $v$ into the magnification equation, we get,$+\frac{1}{3}=\frac{-(-10)}{u}$Simplifying, we get: $u=30\,\text{cm}$Therefore, the object should be placed at a distance of $30\,\text{cm}$ from the spherical mirror to reduce the magnification to $+\frac{1}{3}$.