I'll take a stab at this one. There are some factors to consider prior to estimating the population required for a single state to control both the House and the Electoral Collage. First, the total population (assumed in the original question to be 324,424,000 ppl) is not the population used in determining the number of representatives a particular state has. To get that figure, the populations of all US territories and DC must be removed. For simplicity I'm going to use 2016 population estimates from here. This means that the US population subjected to congressional representation is 322,446,343 ppl.
Another thing on the top that is important, there are 435 members of the House, not the 385 the OP suggests. This means that there are 538 total electors in the Electoral College (435 from the House, 100 from the Senate, and 3 from DC). Our theoretical majority state is still going to need 268 representatives though.
Next, we need to make sure we understand how congressional apportionment works currently. Taking data from the 2010 census, California had a population of 37,254,503 ppl, which lead to 53 seats in the House. Looking at the fraction of the total population living in California,
(37,254,503/308,156,338)*435 = 52.59
This tells us that congressional apportionment rounds up for representation, though there appears to be some considerations required to make total number of voting representatives equal to 435. Here we will consider standard rounding and no funny business to get to 268.
Now let's try to find out the population required for an Electoral College majority. Starting with the total population and the fact that we only need greater than 267.5 effective representatives we can directly calculate the required population:
322446343*(267.5/435) = 198,285,970 ppl (rounded up to the nearest whole person)
This represents approximately 61.5% of the total US population.
The next question you asked was concerning population density. Assuming total state land areas found here, the corresponding population density for California would be 1272.87 pp/mi^2. The lowest possible population density for a state to have a majority of electors in the Electoral College would be Alaska with a population density of 347.48 pp/mi^2.
Following the method of equal proportions referred by @AlonzoMuncy in another answer (which falls under the category of the more you know), it is possible to calculate the population bounds required to a majority of the Electoral College.
In particular, using the priority formula
A = P/Sqrt[n*(n+1)]
it is possible to calculate the upper population bound for our fictitious state directly using the following relationship:
P/Sqrt[n*(n+1)] = (Ptot - P)/Sqrt[(ntot - n)*((ntot - n) + 1)],
P is the population in our state,
Ptot is the total US population,
ntot is the total number of representatives (
ntot = 435), and
n is the number of representatives in our state. Substituting in
n = 268 and
Ptot = 322,446,343 we get that the maximum state population is
P = 198,570,949 ppl
This assumes that all the population is contained in the next populated state. The lower bound can be calculated assuming all other 49 states are equally populated
P/Sqrt[n*(n+1)] = ((Ptot - P)/49)/Sqrt[(ntot - n)/49*((ntot - n)/49 + 1)],
yielding a state population
P = 188,856,488.
This means that our state has to have a population between 188,856,488 and 198,570,949 assuming my understanding of the method is correct.