Solving a parallel RL circuit for the r and l values
Try and find R first. 

You know the applied voltage and you know the power. All that power is dissipated in the resistor so, 69.44 = \$\frac{200^2}{R}\$ hence R = 576.04 ohms.

Can you solve it from here? Hints are not needed any more because the OP spotted his mistake so: -



Knowing apparent power (400 VI) and active power (69.44 watts) you can calculate reactive power as \$\sqrt{400^2-69.44^2}\$ = 393.926 VIr. 

Reactive power is the reactive watts taken by the inductor hence \$X_L\$ is  \$\frac{200^2}{393.926}\$ = 101.542 ohms. Hence L = 323 mH.