AC power!

Let the maximum voltage in the circuit be $V_0$ ,the maximum current be $I_0$ , the angular frequency be $\omega$ and the phase difference in question be $\phi$ .

Assume(without loss of generality):

$V=V_0 \sin {(\omega t)}$

therefore: $I=I_0 \sin {(\omega t+\phi)}$

Hence, instantaneous power is:

$P=VI=\frac {V_0I_0} {2}(\cos {(\phi)}-\cos {(2\omega t+\phi)})$

Hence, maximum instantaneous power is:

$P_{max}=\frac {V_0I_0} {2}(\cos {(\phi)}+1)$ (since $\cos {(2\omega t+\phi)}$ min $=-1$ )

Hence, the average power consumed by the lamp is:

$P_{avg}=<P>=\frac {V_0I_0} {2} \cos {(\phi)}$

It is given that: $P_{avg}=\frac {25} {100} P_{max}$

Hence, we get: $\cos {(\phi)}= \frac {1} {3}$

Therefore: $\boxed{\phi =cos ^{ -1 }{ (\frac { 1 }{ 3 } ) }}$

Note: Common mistake would be to take $P_{max}=V_0I_0$ . This is incorrect as maximum voltage and maximum current may not necessarily occur simultaneously.