\(P =\frac{ V ^2}{ R } \text { so } R =\frac{ V ^2}{ P }\)

Now since the bulb is operating initially at Power \(P_0\) and Voltage \(V_0\). So

\(P_0=\frac{V_0^2}{R}\)

Resistance of bulb is \(R =\frac{ V _0^2}{ P _0}\)

now the bulb is attached to voltage \(V\) and it consumes power \(P\) now. So Using value of \(R\) from equation \((1)\)

\(P=\frac{V^2}{R}=\frac{V^2 P_0}{V_0^2}=\left(\frac{V}{V_0}\right)^2 P_0\)