$\begin{align}  & \frac{1200}{4800}=\frac{1}{{{2}^{n}}A}\Rightarrow {{n}_{A}}=2\xrightarrow[t=12h]{{{n}_{A}}=\frac{t}{({{T}_{\frac{1}{2}}})A}}({{T}_{\frac{1}{2}}})A=6h\,\,(1) \\  & \frac{300}{4800}=\frac{1}{{{2}^{n}}B}\Rightarrow {{n}_{B}}=4\xrightarrow[t=12h]{{{n}_{B}}=\frac{t}{({{T}_{\frac{1}{2}}})B}}({{T}_{\frac{1}{2}}})B=3h\,\,(2) \\  & (1),(2)\Rightarrow \frac{({{T}_{\frac{1}{2}}})A}{({{T}_{\frac{1}{2}}})B}=2 \\ \end{align}$