For the circuit shown in figure, the ammeter $A_2$ reads 1.6A and ammeter $A_3$ reads 0.4A. then

                          

The current of 1.6A lags e.m.f. in phase by$\frac{\pi }{2}$ . The current of 0.4A leads e.m.f. in phase by $\frac{\pi }{2}$ . So, these two currents are ${180}^0$ out of phase with each other.

Net current, $I_1=\left(1.6-0.4\right)A=1.2A$

$X_c=4X_L$

$\Rightarrow \frac{1}{\omega C}=4\omega L$

$\Rightarrow \frac{1}{2\mathrm{\pi f}C}=4\left(2\mathrm{\pi f}\right)L$

$\Rightarrow f=\frac{1}{4\mathrm{\pi }\sqrt{\mathrm{LC}}}$