Ans: D
Since $g(x)$ is divisible by $x+2a$, we have
$\begin{array}{rcl}
g(-2a) & = & 0 \\
(-2a)^2 +a(-2a) +b & = & 0 \\
4a^2 -2a^2 +b & = & 0 \\
b & = & -2a^2
\end{array}$
The required remainder
$\begin{array}{cl}
= & g(2a) \\
= & (2a)^2 +a(2a) +b \\
= & 6a^2 +(-2a^2) \\
= & 4a^2
\end{array}$