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