There is a trick for hex to/from octal because they relate to binary but lets ignore that and consider the general idea (which works for any bases).
Since we want to convert to octal lets start using octal as our number system - as if were always were. So we count like this 1,2,3,4,5,6,7,10,11,12,13,14,15,16,17,20,21,22,...
And our multiplication table is:
$$
\begin{array}{|c|c|c|c|c|c|c|}
\hline
\times & \mathbf 1 & \mathbf 2 & \mathbf 3 & \mathbf 4 & \mathbf 5 & \mathbf 6 & \mathbf 7\\
\hline
\mathbf 1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\mathbf 2 & 2 & 4 & 6 & 10 & 12 & 14 & 16 \\
\mathbf 3 & 3 & 6 & 11 & 14 & 17 & 22 & 25 \\
\mathbf 4 & 4 & 10 & 14 & 20 & 24 & 30 & 34 \\
\mathbf 5 & 5 & 12 & 17 & 24 & 31 & 36 & 43 \\
\mathbf 6 & 6 & 14 & 22 & 30 & 36 & 44 & 52\\
\mathbf 7 & 7 & 16 & 25 & 34 & 43 & 52 & 61 \\
\hline
\end{array}
$$
So the hexadecimal digits are $0\text x1 = 1$, $0\text x2 = 2$, $0\text x3 = 3$, $0\text x4 = 4$, $0\text x5 = 5$, $0\text x6 = 6$, $0\text x7 = 7$, $0\text x8 = 10$, $0\text x9 = 11$, $0\text {xA} = 12$, $0\text xB = 13$, $0\text xC = 14$, $0\text xD = 15$, $0\text xE = 16$, $0\text xF = 17$.
Anyway, we want to re-interpret the number 0x1A03
which is equal to $ 20^3 \cdot 1 + 20^2 \cdot 12 + 20 \cdot 0 + 3$. Now we just "do the math", multiply everything out and add it up.
$$
\begin{align}
& 20^3 \cdot 1 + 20^2 \cdot 12 + 20 \cdot 0 + 3 \\
=& 10000 + 400 \cdot 12 + 3 \\
=& 10000 + 5000 + 3 \\
=& 10000 + 5000 + 3 \\
=& 15003
\end{align}
$$
Now the fact that we have all these zeros suggests there would be a quicker method for these particular bases (normally we would not get so many zeros in our calculations).