$\begin{array}{l} \\ r_1=radius of outer sphere,r_2=radius of inner sphere\\ V_1=Volume of outer sphere=\frac{4{{\mathrm{\pi r}}_1}^3}{3}\\ ,V_2=Volume of outer sphere=\frac{4{{\mathrm{\pi r}}_2}^3}{3}\\ Volume of hollow sphere=V_1-V_2\\ Rate of increase in radius of outer sphere=v cm/sec\\ \mathrm{V}=\frac{4\mathrm{\pi }}{3}\left({\mathrm{r}}_1^3-{\mathrm{r}}_2^3\right)\Rightarrow 3{\mathrm{r}}_1^2\frac{{\mathrm{dr}}_1}{\mathrm{dt}}-3{\mathrm{r}}_2^2\frac{{\mathrm{dr}}_2}{\mathrm{dt}}=0\\ \frac{{\mathrm{dr}}_2}{\mathrm{dt}}=\frac{{\mathrm{r}}_1^2}{{\mathrm{r}}_2^2}\frac{{\mathrm{dr}}_1}{\mathrm{dt}} \\ given that r_1=a+d,r_2=a\\ \Rightarrow \frac{{\mathrm{dr}}_2}{\mathrm{dt}}=\frac{(\mathrm{a}+\mathrm{d}{)}^2}{{\mathrm{a}}^2}v cm/sec\end{array}$