We also know that $\mathbb P(A\cap B\cap C)=0.001\%=10^{-5}$. What we need to calculate is $\mathbb P(C|B\cap A)$. \begin{align*} \mathbb P(C|B\cap A)&=\frac{\mathbb P(A\cap B\cap C)}{\mathbb P(B\cap A)}\\\&=\frac{\mathbb P(A\cap B\cap C)}{\mathbb P(A)}\cdot\frac{\mathbb P(A)}{\mathbb P(B\cap A)}\\\&=\frac{\mathbb P(A\cap B\cap C)}{\mathbb P(A)}\cdot\frac{1}{\mathbb P(B|A)}\\\&=\frac{10^{-5}}{10^{-1}}\cdot10^2\\\&=10^{-2}=1\%. \end{align*}