$1+2+2^2+2^3+....+2^n>9999$
$\implies\frac{2^{n+1} -1}{2-1} >9999$ (sum of GP series)
$\implies 2^{n+1} >10000$
Taking log on both sides
$\implies ({n+1})*log_{10}2> 4*log_{10}10$
$\implies (n+1)*0.30103 > 4$ (given $log_{10}2=0.30103$)
$\implies n + 1 > \frac{4}{0.30103}$
$\implies n +1 > 13.29$
$\implies n > 12.29$
$\therefore$ Option $B.$ is the correct answer.