Capacitive reactance is $X_C=\frac{1}{\omega C}=\frac{1}{3000\times 25\times {10}^{-6}}=\frac{40}{3}\mathrm{\Omega }$

Inductive reactance is $X_L=\omega L=3000\times 10\times {10}^{-3}=30\mathrm{\Omega }$

Now, total reactance is $X=X_L-X_C$

$=30-\frac{40}{3}=\frac{50}{3}\mathrm{\Omega }$

$\tan \varphi =\frac{X}{R}=\frac{{\displaystyle \frac{50}{3}}}{100}=\frac{1}{6}$

$\tan \varphi =\frac{1}{6}=0.17$

Hence, the phase difference between the applied voltage and resultant current is ${\tan }^{-1}\left(0.17\right)$.