Say, Anne can finish 'a' amount of work in 1 hour, and bruce can finish 'b' amount of work in 1 hour. If they worked together at their constant rates, they will do 'a+b' amount of work in 1 hour .
Therefore, let W be the total amount of work done in cleaning the house.
If the W amount of work done in 4 hours, w= 4(a+b)
if Annes rate is doubled, she can do (2a) amount of work in 1 hour.
for three hours, $$w=3\left(2a+b\right)\ \left(anne\ doubled\right)$$
$$i.e\ 4\left(a+b\right)=3\left(2a+b\right)\ $$
$$4b-3b=6a-4a$$
$$or\ b=2a$$
$$i.e\ w=4\left(a+2a\right)=12a$$
$$If\ anne\ do\ the\ whole\ work\ \left(i.e\ 12a\right)\ at\ her\ normal\ rate\ \left('a'\ amount\ of\ work\ in\ 1\ hour\right)$$
$$Then\ the\ number\ of\ hours\ used=\frac{\left('12a'work\right)}{'a'\ \frac{work}{hour}}=12hours$$
$$Hence,\ option\D\ is\ very\ correct$$