Answer
$112.6^\circ\rm C$
Work Step by Step
Since we are dealing with emitting energy by radiation, so we can use the power formula of
$$P_{w1}=\left(\dfrac{Q}{t}\right)_{1}=\epsilon \sigma A\left[T_1^4-T_{room}^4\right]$$
Where 1 refers to the first lightbulb of 75-W.
And since $90\%$ of energy emitted is heat, so
$$0.9P_{w1} =\epsilon \sigma A\left[T_1^4-T_{room}^4\right]\tag 1$$
Solving for $\epsilon \sigma A$ for the bulb.
$$\epsilon \sigma A=\dfrac{0.9P_{w1}}{ \left[T_1^4-T_{room}^4\right]}$$
Plugging the known and remember to convert the units of temperatures to kelvin.
$$\epsilon \sigma A=\dfrac{0.9\cdot 75}{ \left[348^4-291^4\right]}$$
$$\epsilon \sigma A=9.0056\times10^{-9}\;\rm W/K^4\tag 2$$
Using equation (1) for the second bulb to solve for $T_2$;
$$0.9P_{w2} =\epsilon \sigma A\left[T_2^4-T_{room}^4\right] $$
$$\dfrac{0.9P_{w2}}{\epsilon \sigma A} = T_2^4-T_{room}^4 $$
$$T_2^4=\dfrac{0.9P_{w2}}{\epsilon \sigma A} +T_{room}^4 $$
$$T_2 =\sqrt[4]{\dfrac{0.9P_{w2}}{\epsilon \sigma A} +T_{room}^4 }$$
Plugging the known and from (2);
$$T_2 =\sqrt[4]{\dfrac{0.9\cdot 150}{9.0056\times10^{-9}} +291^4}=\bf 385.8\;\rm K$$
Therefore,
$$T_2=\bf112.6^\circ\rm C$$
