For bulb $1$

$I=\frac{V}{P}=\frac{60}{200}=\frac{6}{20} A$

Hence,

$R=\frac{V}{I}=\frac{200 \times 20}{6}=\frac{4000}{6} \Omega$

Similarly, for bulb $2$

$I=\frac{V}{P}=\frac{100}{200}=\frac{1}{2} A$

Hence,

$R=\frac{V}{I}=\frac{200 \times 2}{1}=400 \Omega$

When bulbs are connected in series their resistances will added up

$R^{\prime}=\left(\frac{4000}{6}\right)+400$

$R^{\prime}=\frac{6400}{6}$

Hence, current is

$I^{\prime}=\frac{V}{R^{\prime}}=\frac{200 \times 6}{6400}=\frac{12}{64} A$

Hence, $P^{\prime}=V I=200 \times \frac{12}{64}$

$P^{\prime}=37.5 W$