Answer
$^{30}Si$
Work Step by Step
Recall: The radius of nucleus $R=r_{0}A^{1/3}$
where $r_{0}=1.2\,fm$ and $A$ is the mass number.
Given $2R=7.46\,fm\implies R= \frac{7.46}{2}\,fm=3.73\,fm$
$A=(\frac{R}{r_{0}})^{3}=(\frac{3.73\,fm}{1.2\,fm})^{3}=30$.
A stable nucleus with mass number $30$ is $^{30}Si$.
