Explanation:

A Given,
\(\mathrm{E}=9 \times 10^{13} \mathrm{~J}\)
\(\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-2}\)
We know that,
\(\mathrm{E}=\mathrm{mc}^{2}\)
\(9 \times 10^{13}=\mathrm{m}\left(3 \times 10^{8}\right)^{2}\)
\(\mathrm{~m}=\frac{9 \times 10^{13}}{\left(3 \times 10^{8}\right)^{2}}=\frac{9 \times 10^{13}}{9 \times 10^{16}}\)
\(\mathrm{~m}=1 \times 10^{-3} \mathrm{~kg}\)
\(\frac{\Delta \mathrm{m}}{\mathrm{m}} \times 100=10^{-3} \times 100=0.1 \%\)