I see what you did there in 2016

Let $u=2^{2016}$ and $v=3^{2016}$ :

$\sqrt{4^{2016}+2 \times 6^{2016}+9^{2016}}=\sqrt{\left (2^{2} \right)^{2016}+ 2 \times \left (2 \times 3 \right)^{2016}+\left (3^{2} \right)^{2016}}=\sqrt{u^2+2uv+v^2}$

$\cdots=\sqrt{(u+v)^2}=u+v=2^{2016}+3^{2016}$

So the problem is reduced to finiding $2^{2016}$ and $3^{2016} \pmod{1000}$ or equivalently $\pmod{8}$ and $\pmod{125}$ . We will use $\phi (125)=100$ :

$\begin{cases} 2^{2016} \equiv 8 \times 2^{2013} \equiv 0 & \pmod{8} \\ 2^{2016} \equiv 2^{16} \equiv \left (2^8 \right)^2 \equiv 6^2 \equiv 36 & \pmod{125} \end{cases} \implies 2^{2016} \equiv 536 \pmod{1000}$

$\begin{cases} 3^{2016} \equiv 9^{2013} \equiv 1 & \pmod{8} \\ 3^{2016} \equiv 3^{16} \equiv \left (3^8 \right)^2 \equiv 61^2 \equiv 96 & \pmod{125} \end{cases} \implies 3^{2016} \equiv 721 \pmod{1000}$

$2^{2016}+3^{2016} \equiv 536+721 \equiv \boxed{\boxed{257}} \pmod{1000}$