\(\mathrm{f}=1 \mathrm{\,MHz}=10^{6} \mathrm{\,Hz}\)
\(\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)
\({{\rm{f}}^2} = \frac{1}{{4{\pi ^2}{\rm{LC}}}}\)
\( \Rightarrow C = \frac{1}{{4{\pi ^2}{{\rm{f}}^2}L}} = \frac{1}{{4 \times 10 \times {{10}^{ - 2}} \times {{10}^{12}}}}\)
\(=\frac{10^{-12}}{4}=2.5 \mathrm{\,pF}\)