To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$. To Find $a$ such that  $2^{1990} \equiv a\pmod {1990}$.
$1990 = 2 \times 5 \times 199$. Now $a \equiv 0 \pmod {2}$, $a \equiv 4 \pmod{5}$ and $a \equiv 29 \pmod{199}$. Taking first two together we get $a \equiv 4 \pmod {10}$.
$199x \equiv 1\pmod {10}$ has a solution $x =-1$ and $10x \equiv 1\pmod{199}$ has a solution $x = 20$.
Thus using Chinese Remainder Theorem we have, 
$a \equiv 4(-1)199 + 29(20)10 \equiv 5004 \equiv 1024\pmod {1990}$.
Thus our $a = 1024$.
Is the solution correct? Is there any better proof?

Checking your answer of $a=1024=2^{10}$: 
$$\begin{align}2^{1990}&\equiv 0\equiv 2^{10}&\pmod 2\\
2^{1990}&\equiv \left(2^{4}\right)^{495}2^{10}\equiv 2^{10}&\pmod{5}\\
2^{1990}&\equiv \left(2^{198}\right)^{10}2^{10}\equiv 2^{10}&\pmod{199}
\end{align}$$
After your edits, your answer is correct.
