d
Nuclear radii \(R=\left(R_{0}\right) A^{1 / 3}\)

where \(A\) is the mass number.

\(\therefore \frac{{{R_{{\text{Ge}}}}}}{{{R_{{\text{Al}}}}}} = {\left( {\frac{{{A_{{\text{Ge}}}}}}{{{A_{{\text{Al}}}}}}} \right)^{1/3}}\) \( = {\left( {\frac{{125}}{{27}}} \right)^{1/3}} = \left( {\frac{5}{3}} \right)\)

or, \(R_{Ge}=\frac{5}{3} \times R_{A l}=\frac{5}{3} \times 3.6=6 \mathrm{\,fm}\)

\((Given \;R_{\mathrm{Al}}=3.6 \mathrm{\,fm})\)