Yay for 2014 in 2015

Factor $2015^3$ out from our desired value. We have

$\begin{aligned} (201502015.02015)^3 &= 2015^3 \times (100001.00001)^3 \\ &= 2015^3 \times (10^5 + 1 + 10^{-5})^3 \end{aligned}$ .

Expanding $(10^5 + 1 + 10^{-5})^3$ using $(a + b + c)^3 = a^3 + b^3 + c^3 + 3(a^3b + ab^2 + a^2c + ac^2 + b^2c + bc^2) + 6abc$ gives

$(100001.00001)^3 = 10^{15} + 3(10^{10}) + 6(10^5) + 7 + 6(10^{-5}) + 3(10^{-10}) + 10^{-15}.$

Finally, multiply $2015^3$ , which we know the value of, by these powers of 10. Careful arithmetic gives

$(201502015.02015)^3 = 8181598820510119294964508.656914193853375,$

which has digit sum $\boxed{189}.$

First I calculate a floating point approximation to the value, just to see what I'm aiming for. I split the item to be cubed into its whole and decimal parts and represent each as a fraction. Python's fractions module calculates with fractions, even when the numbers become very large. Now I calculate the required result as a fraction. Since there is only one leading digit in the denominator, an 8, I can multiply the numerator by 125 to obtain a representation of the numerator of the result (with the thought in mind that multiplying the denominator by this same number would 'clear' the 8 leaving only a one in the leftmost place).

Then I simply add the digits.