Additional buoyant force 
$\begin{aligned}
& g \rho a^2 x=\sigma a^3 A \
& A=\frac{\rho}{\sigma} \frac{g}{a} x \
& T=2 \pi \sqrt{\frac{\sigma a}{\rho g}}
\end{aligned}Now,\sigma=\frac{10 \times 10^{-3}}{10^{-3}}=10\begin{aligned} \Rightarrow \quad T & =2 \pi \sqrt{\frac{10 \times 0.1}{10^3 \times 10}} \ & =2 \pi \times 10^{-2}\end{aligned}$